Chapter-7, COORDINATE GEOMETRY

Explanation:

By assuming a point (2, 0) on the x-axis, it is possible to calculate the distance between the points (2, 3), and the x-axis.

x2-x12 + y2-y12 is the distance between x1, y1, and x2, y2.

The distance between (2, 3) and (0, 2) is, therefore, 2-22+0, 32 = 32 = 9 = 3.

As a result, the point (2, 3) is 3 units away from the x-axis.

Explanation:

We are aware that the distance between two points A(x1, y1) and B(x₂, y₂) may be calculated using the formula Distance=√(x₂- x1)2+(y₂−y1)2(x₂−x1)2+(y₂−y1)2

Finding the distance between points A (0, 6) and B (0, -2) is given.

When these values are substituted in the equation, we get

Distance equals √(0- 0)2+(−2−6)2(0−0)2+(−2−6)2

hence, we do

√64 miles separate A and B.

A and B are separated by 8 units.

As a result, there are 8 units between points A and B.

Explanation:

We are aware that Distance = (x₂-x1) can be used to calculate the distance between two places P(x1, y1) and Q(x₂, y₂)2+(y₂−y1)2(x₂−x1)2+(y₂−y1)2

Given that, we must determine the amount of time between point P (-6, 8) and the origin (0, 0),

These variables are substituted in the equation to provide Distance = (0, 6)2+(0−8)2(0−(−6))2+(0−8)2

As a result, we arrive at the Distance of point P = 36 + 64 = 100 = 10.

Point P is 10 units away.

As a result, the point P (-6, 8) is 10 units away from the origin.

Explanation:

We are aware that Distance = (x₂-x1) can be used to calculate the distance between two places P(x1, y1) and Q(x₂- y₂)2+(y₂−y1)2(x₂−x1)2+(y₂−y1)2

Finding the distance between the points √(0, 5) and (-5, 0) is given.

When these values are substituted in the equation, we get Distance=(-5-0)2+(0-5)2

As a result, we arrive at Distance between the Points =√ (25 + 25) = √50 = 5√2.

the points are separated by 5√2 units.

As a result, it is 5√2 units between the positions (0, 5) and (-5, 0).

Explanation:

Because AOBC is a square with the vertices A (0, 3), O (0, 0), and B (5, 0) as its three corners,

The distance between the points (0,3) and (5,0) is the diagonal's length (AB).

We are aware that the distance between two places P(x1, y1) and Q(x₂, y₂) may be calculated using the formula Distance=√(x₂-x1)2+(y₂−y1)(x₂−x1)2+(y₂−y1)2

Distance=(50) can be used to calculate the distance between the points (0,3) and √(5-0)2+(0−3)2=(5−0)2+(0−3)2

As a result, we arrive at Distance between the Points = √(25 + 9) = √34.

34 units separate the spots in their distance.

As a result, its diagonal is √ 34 units long.

Explanation:

the length of each side of the first.

Make A(4, 0), B(0, 0), and C (0, 3) the triangle's vertices.

According to the distance calculation, √AB = 42 + 0 = 4, BC = √0 + 32 = 3, and CA = √42 + 32 = 5.

It is evident that combining AB + BC + CA results in the perimeter of ABC equals 3+4+5=12 units.

Explanation:

The triangle's vertices are marked as A (3, 0), B (7, 0), and C (8, 4).

The equation to determine a triangle's surface area is

Area is equal to half of [x1(y₂–y3) + x₂(y3–y1) + x3(y1–y₂)].

Changing the values

Area = 1/2 [3(0 - 4) + 7(4 - 0) + 8(0 - 0)]

Area may be calculated as Area = [3(-4) + 7(4) + 8(0)]/[1/2].

Area = 1/2 [-12 + 28]

The result is Area = 1/2 [16].

8 square units of area

Consequently, a triangle has an area of 8 square units.

Explanation:

Provide the vertices A(-4,0), B(4,0), and C(0,3).

As for the separation between A(-4,0) and B(4,0),

∵ The separation between (x1,y1) and (x₂,y₂)

D=√(x₂−x1)2+(y₂−y1)2 AB=√[4−(−4)]2+(0−0)2]

=√(4+4)2=√82=8

The separation between B(4,0) and C(0,3), where BC=√(0-4)2+(3−0)2=√16+9=√22=5

Distance√ [0,(4)] between A(-4,0) and C(0,3)2+(3−0)2=√16+9=√25=5

∴ BC=AC

Because an isosceles triangle has two equal sides, ABC is an isosceles triangle.

Explanation: Ans

We are aware that the parameters of the point P are [m1x₂+m₂x1/m1+m₂,m1y₂+m₂y1/m1+m₂], which internally splits the line segment joining the points A(x1, y1) and B(x₂, y₂) in the ratio m1: m₂.

(X1, Y1) = (7, -6) (X₂, Y₂) = (3, 4) m1 = 1 m₂ = 2 are supplied.

The values in the section formula, [13+271+2,14+261+2]=(173,83), are substituted.[1×3+2×7/1+2,1×4+2×−6/1+2]=(17/3,−8/3)

due to having a positive x location and the inverse y value

The IV quadrant contains the point.

As a result, the points (7, -6) and (3, 4) in the internal ratio of 1: 2 are located in the IV quadrant.

Explanation:

We are aware that the midpoint of the line segment connecting the points P (x₁, y₁) and Q (x₂, y₂) is located at [(x₁ + x₂)/2, (y₁ + y₂)/2].

The scores are as follows: A (-2, -5) and B (2, 5).

The values in the formula = [(-2 + 2)/2, (-5 + 5)/2] can now be changed.

= (0/2, 0/2) = (0, 0)

As a result, (0, 0) is the position where the perpendicular is located.

Explanation:

Let D(x,y) in a parallelogram bisect each other, in which case the midpoints of diagonals AC and BD are calculated using the centre of the formula.

((2+8) / 2 = (6+x) / 2 (using the x coordinates) ((2+8) / 2 = (3+3) / 2) ((6+x) / 2, (7+y) / 2)

D (0,1) is the result of equation 3 =(6+x)/2

6 = 6+x

x = 0 and y = 1.

Explanation:

Let D(x,y) in a parallelogram bisect each other, in which case the midpoints of diagonals AC and BD are calculated using the center of the method.

((2+8) / 2 = (6+x) / 2 (using the x coordinates) ((2+8) / 2 = (3+3) / 2) ((6+x) / 2, (7+y) / 2)

D (0,1) is the result of equation 3 =(6+x)/2

6 = 6+x

x = 0 and y = 1.

Explanation:

P is the center of the line section QR Px = a/3 = (-6 - 2)/2, which is known.

Cross multiplication yields the result 2a = -24.

2 divided by both sides yields a = -12.

As a result, a has a value of -12.

Explanation:

A(x₁,y₁)=(1,5) and B(x₂,y₂)=(4,6) are the supplied locations.

The center of AB will be traversed by the perpendicular bisector of AB.

At P(0,y), the parallel occurs despite AB will now intersect the Y axis.

∴AP=BP

By using the distance formula, we obtain AP2=BP2: (x₁-0)2+(y₁-y)2=(x₂-0-y)2.2 ⇒(1−0)2+(5−y)2=(4−0)2+(6−y)2

When we solve, we obtain y=13.

The point is therefore (0,13).

Explanation:

Think of P (h, k) as the position of the point that is equally distant from the three vertices O (0, 0), A (0, 2y), and B (2x, 0).

Here PA = PB + PO

(1) PO2 = PA2 = PB2

[√(h−0)2+(k−0)2]2=[√(h−0)2+(k−2y)2]2=[√(h−2x)2(k−0)2]2[(ℎ−0)2

The result is now h2 + k2 = h2 + (k - 2y)2 = (h - 2x)2 + k2 --- (1)

Let's have a look at the first two equations.

h2 + k2 = h2 + (k - 2y)2 k2 = k2 + 4y2 - 4yk

Leaving out the everyday language

Where y 0, 4y(y - k) = 0 and y = k

H2 + K2 = (h - 2x) when the first and third equations are taken into account.H2 + 4x2 - 4xh = 2 + k2 h2

Leaving out the everyday language

Where x 0, 4x(x - h) = 0 and x = h

Therefore, the necessary points are (h, k) = (x, y).

As a result, the point's coordinates are (x, y).

Explanation:

It is assumed that a circle with an origin at (0, 0) and passing through (13/2, 0) is drawn.

The fact that

The radius of the circle equals the distance between the two points, or (13/2 0)2+(0 0)2=(132)2=132=6.5.

If a point's distance from the circle's center is less than, equal to, or more than the radius of the circle, the point is either within, on, or outside the circle.

The distance between (0, 0) and (-3/4, 1) is (34- 0)2 + (1- 0)2 = 916+1 = 2516 = 5/4 = =1.25<  6.5 (-3/4, 1), which is located in the center of the circle. Let's now examine each choice one at a time.

b. The distance between (0, 0) and (2, 7/3) is equal to ((2 0)2+(73 0)2=4+499=36+499=859) = 9.22/3 = 3.1 6.5 (2, 7/3) and is located inside the circle.

The separation between (0, 0) and(5, -1/2) = √( 5 -0 )2 + ( 12- 0 )2 = √25 + 14 =√ 1014 = ( 5 -0 )2 + ( 12- 0 )2 

= 10.04/2 

= 5.02 6.5 (5, -1/2) is located inside the circle.

d. The point on the circle where (0, 0) and (-6, 5/2) are separated by √(-6-0, 52, 36, 144+ 254+ and 1694 is (6-0) 52, 36, 144, 254, and 1694 = 13/2, or 6.5).

The point (-6, 5/2) does not, therefore, lie within the circle's interior.

Explanation:

Let P's and Q's coordinates be (x,y) and (x₂,y₂), respectively.

The center of PQ = (2, 5)

Using a halfway solution

Y=y₁+y₂/2 and x=x₁+x₂/2

2 = x₁ + x₂/2 and 5 = y₁ + y₂/2

Y₁+Y₂ = 10 and x₁+x₂ = 4.

Given that line, PQ crosses the y-axis at P

So, x₁=0. 

Likewise, y₂=0

∴x2=4 and y₁=−10

P's values are (0,10) and Q's values are (4,0).

Explanation:

Area of triangle ABC is known to be equal to 1/2 [x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)].

Let's insert [a (c + a - a - b) + b (a + b - b - c) + c (b + c - c - a)] with the values = 1/2.

Simplifying even further,Δ  = 1/2 [a (c - b) + b (a - c) + c (b - a)]

Δ[ac - ab + ab - bc + bc - ac] = 1/2

We thus obtain Δ= 1/2 [0].

Δ = 0

Consequently, the shape's surface is equal to zero.

Explanation:
The right response is B 0.

Triangle's area is equal to 1/2 [x₁(y₂ - y₃) + x₂ (y₃ - y₁ + x₃ (y₁ - y₂)].

In other words, 12[ac - ab + ab - bc + bc - ac] 

= 12[ac - ab + ab - bc + bc - ac] = 0.

Explanation:

The right answer is C 2a = b.

Let A = (x₁,y₁)=(1,2) B = (x₂,y₂)=(0,0) and C = (x₃,y₃)=(a,b) be the supplied points.

Area of ABC is equal to 1/2[x₁(y₂y₃)+x₂(y₃y₁)+x₃(y₁y₂)].

∴Δ=1/2[1(0−b)+0(b−2)+a(2−0)]

=1/2(−b+0+2a)=12(2a−b)

Due to the collinearity of points A(1,2), B(0,0), and C(a,b), the triangle ABC's area should be equal to zero.

Area of ABC = 0 (12(2a + b) = 0(2a + b) = 0(2a + b)

The necessary relation is therefore 2a = b.

Explanation:

The vertices of the triangles A(-2,0), B(2,0), C(0,2), D(-4,0), E(4,0), and F(0,4) are known.

Units AB=(0+2)2+(20)2 =4+4=22.

BC = ((20)2 + ((02)2) = 4 + 4 = 22 units

CA=(−2−2)2+(0−0)2 =(−4)2+0=4 units 

FD=(0+4)2+(0−4)2 =42+(−4)2=42 units

FE=(4−0)2+(0−4)2 =42+(−4)2=42 units 

DE = ((4)(2)+(0)(2) = ((8)2=(8) units.

In this case, we can see that sides of DEF are twice as big as the sides of ABC.

Therefore, both triangles are comparable.

Explanation:

Determine if the point P(-4, 2) is on the line segment given that it connects points A(-4, 6) and B(-4, -6).

Q (x₁, y₁) is connected by the line segment y - y₁ = m(x - x₁).

where the slope m is given.

The equation for the slope of the line connecting the two points P (x₁, y₁) and Q (x₂, y₂) is m = (y₂ - y₁)/(x₂ - x₁).

A and B (in this case, -4, 6)

Slope, m = 6 - 6/(4 - (-4) = 12 / 0 =

The line segment now has the equation: y - 6 = (12/0)(x - (-4))(y - 6)(0) = -12(x + 4)

0 = -12(x + 4) x + 4 = 0 -------------- (1)

The point P(-4, 2) must satisfy equation (1) if it is on the line segment.

Add x = -4 to (1).

Explanation: 

The positions are as follows: (0, 5), (0, -9)), and (3, 6).

We must determine whether the locations are collinear.

A triangle's area is 1/2[x₁(y₂ - y₃) + x2(y₃ - y₁) + x₃(y₁ - y₂)] when its vertices are A (x₁, y₁), B (x₂, y₂), and C (x₃, y₃).

The triangle's area must be zero in order to determine whether the points are collinear.

In this instance, (x₁, Y₁) = (0, 5), (x₂, Y₂) = (0, -9),

and (x₃, Y₃) = (3, 6)

Triangle's area is equal to 1/2 [0(-9 - 6) + 0(6 - (-9)) + 3(5 - (-9)] 

= 1/2 [0 + 0 + 3(5 + 9)]

= 1/2 [3(14)].

= 3(7)

 = 21

Triangle area = 0

As a result, the specified points are not parallel.

Explanation: 

Determine whether the point P(0, 2) is the point of intersection of the provided line section's vertical bisector and the y-axis given a line segment connecting the points A(-1, 1) and B(3, 3).

We are aware that any point located on the line segment's perpendicular bisector that connects two points is equally far from those two positions.

A(-1, 1) and B(3, 3) make up the perpendicular bisector connecting the two sites.

If the point P is on the perpendicular bisector, then the distance between PA and PB, or PA = PB, must be equal.

√[(x₂ - x₁)²+(y₂ - y₁)²] is the distance between two points P (x₁, y₁) and Q (x₂, y₂).

To calculate the separation between P(0, 2) and A(-1, 1) PA= √[(-1 - 0)²+(1 - 2)²] = √[(-1)²+(-1)²] = √(1+1) = √2

Determine the separation between P(0, 2) and B(3, 3).

PB = √[(3 - 0)²+(3 - 2)²] = √[(3)²+(1)²] = √(9+1) = √10

It's obvious that PA ≠ PB.

Location P(0, 2) cannot, thus, sit on the line clip's perpendicular bisector.

Explanation: 

If the sum of any two sides is greater than the third of each other then locations A, B, and C may create a triangle. 

AB=310 units since AB²=(x₂-x₁)²+(y₂-y₁)² 

and AB²=(12-3)²+(2-1)=81+9=90. 

BC2=(0−12)²+[2−(−2)]² 

= 144 + 16 = 160; BC = 410 units. 

AC2=(0-3)²+(2-1)²9+1=10, where AC = 10 units and AB = 310 units. 

At this point, AB+AC=10+310=410 units = BC. 

A, B, and C points cannot, therefore, make A.

Explanation:

The points are A(4, 3), B(6, 4), C(5, -6), and D(-3, 5), as shown.

The given points must be examined to see if they are parallelogram vertices.

The opposite sides of a parallelogram are equal by virtue of the parallelogram's attribute.

[(x₂ - x₁)] is the distance between two places P (x₁- y₁) and Q (x₂,- y₂).²+(y₂ - y₁)²]

AB = √[(6 - 4)²+(4 - 3)²]

= √[(2)²+(1)²]

= √(4 + 1)

AB = √5 BC = √[(5 - 6)²+(-6 - 4)²]

= √[(-1)²+(-10)²]

= √(1 + 100)

BC = √101

CD = √[(-3 - 5)²+(5 - (-6))²]

= √[(-8)²+(11)²]

= √(64 + 121)

CD = √185 AD = √[(-3 - 4)²+(5 - 3)²]

= √[(-7)²+(2)²] = √(49 + 4)

AD = √53

There is no doubt that AB CD and BC-AD.

They are not equal, the opposing factions.

As a result, the specified locations are not parallelogram's vertices.

Explanation: 

Given that, the circle's middle is at its origin.

P(5, 0) is a point on the circle.

Whether or not the location Q(6, 8) is outside the circle must be determined.

We know that if any point's distance from its central location is less than or equal to the radius, the point is within the circle, and if it is greater than or equal to the radius, the point is in the circle.

• exceeds the radius, in which case the point is outside the sphere.

To determine the circle's radius,

The circle's radius is determined by the distance between its origin and point P(5, 0).

√[(x₂ - x₁)²+(y₂ - y₁)²] is the distance between two points P (x₁, y₁) and Q (x₂, y₂).

OP = radius of the circle hence equals [(5- 0)²+(0 - 0)²]

= √[(5)²+0]

Length from the origin to Q(6, 8): = 25 OP = 5 = √[(6 - 0)²+(8 - 0)²]

= √[(6)²+(8)²]

= √(36 + 64) 

= √100 

OQ = 10

We see that OQ exceeds OP.

The radius of the circle is bigger than the distance to the point OQ.

The point Q is therefore outside the circle.

Explanation:

If the line segment connecting the coordinates P(6, 5) and Q(0, -4) is known, we must ascertain whether point A(2, 7) is located on the line segment's perpendicular bisector.

We are aware that any point located on the line segment's perpendicular bisector that connects two points is equally far from those two positions.

P(6, 5) and Q(0, -4) are the perpendicular bisectors that connect the two points.

Length from PA and QA must be equal in order to determine if point A is on the perpendicular bisector.

Thus, PA = QA

[(x₂- x₁)² + (y₂ - y₁)²] is the distance between two places P (x₁, y₁) and Q (x₂, y₂).

The gap from P(6, 5) and A(2, 7) is calculated as [(2 - 6)² + (7 - 5)²] = [(-4)² + (2)²].

= √(16 + 4) = √20 = 2√5

The separation from Q(0, -4) and A(2, 7) = √[(2 - 0)² + (7 - (-4))²]

= √[(2)² + (11)²]

= √(4 + 121)

 = √125 

= 5√5

PA ≠ QA

As a result, location A(2, 7) is not on the section of the line that bisects the supplied locations and is perpendicular to it.

Explanation: 

If the line section connecting points A(7, -2) and B(1, -5) is supplied, we must determine if point P(5, -3) constitutes one of the two locations of trisection of the given line segment.

Using the section formula

[(kx₂ + x₁)/k + 1, (ky₂ + y₁)/k + 1] are the coordinates of the point P(x, y), which internally splits the line segment between the points A (x₁, y₁) and B (x₂, y₂) in the ratio k: 1.

The line segments A(7, -2) and B(1, -5) are internally divided by P(5, -3) in the ratio k:1.

Thus, (5, -3) = [k(1) + (7)]/k + 1 and (k(-5) + (-2))/k + 1 respectively.

(5, -3) = [k + 7/k + 1, -

5k - 2/k + 1]

Since k + 7/k + 1 = 5,

k + 7 equals 5(k + 1).

k + 7 = 5k + 5

By arrangement, 7 - 5 5k k

4k = 2 

k = 2/4

 k = 1/2

The line segment AB is divided in half at position P(5, -3) in the ratio 1:2.

As a result, the location P is where AB trisects.

Explanation:

The points are A(-6, 10) given. B(-4, 6) and C(3, -8)

We must determine whether the points are parallel and demonstrate that AB = 2/9 AC.

A triangle's area is 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)] when its vertices are A (x₁, y₁), B (x₂, y₂), and C (x₃, y₃).

The triangle's area must be zero in order to determine whether the points are collinear.

In this case, (x₁, y₁) = (-6, 10), (x₂, y₂) = (-4, 6), and (x₃, y₃) = (3, -8)

Triangle's area is equal to 1/2 [-6(6 - (-8)) + -4(-8 - 10) + 3(10 - 6)] + 1/4 [-6(14) + -4(-18) + 3(4)].

= 1/2[-84 + 72 + 12]

= 1/2[-84 + 84]

= 0

Triangle's surface area is zero.

The points are thus parallel.

the separation of two points P (x1, y1) and

Q (x₂,y₂) is [(x₂, x₁)² + (y₂, y₁)]²]

The separation between A(-6, 10) and B(-4, 6)

AB = √[(-4 - (-6))² + (6 - 10)²]

= √[(2)² + (-4)²]

= √(4 + 16) 

= √20 

= 2√5

The separation between A(-6, 10) and C(3, -8) AC = √[(3 - (-6))² + (-8 - 10)²]

= √[(9)² + (-18)²]

= √(81 + 324) = √405 = 9√5

As a result, 2/9 AC = 2/9(95) = 25 = AB.

Consequently, AB = 2/9 AC.

Explanation:

False

We say that an area is on a circle if the distance between it and the center is equal to the radius.

Now, the distance between P(-2,4) and the center (3,5) equals (x₂-x₁)²+(x₂-y₁)², which is not the same as the circle's radius, and the distance between the points (x₁,y₁) and (x₂,y₂) equals (x₂-x₁)²+(x₂-y₁)². As a result, the circle does not pass through the location P(-2,4).

Explanation:

The points are A(-1, -2) provided. B(4,3), C(2,5), and D(-3,0)

The provided points must be examined to see if they can be arranged to form a rectangle.

We are aware that a rectangle's diagonals are equal.

[(x₂ - x₁)² + (y₂ - y₁)] is the distance between two places P (x₁, y₁) and Q (x₂, y₂).²]

A(-1, -2)'s separation from B(4, 3) = √[(4 - (-1))² + (3 - (-2))²]

= √[(5)² + (5)²]

= √(25 + 25)

= √50 AB = 5√2

Distance [(2 - 4)² + (5 - 3)²] between B(4, 3) and C(2, 5)

= √[(-2)² + (2)²]

= √(4 + 4) = √8 BC = 2√2

Distance [(-3 - 2)²+ (0 - 5)²] between C(2, 5) and D(-3, 0).

= √[(-5)² + (-5)²]

= √(25 + 25) = √50 CD = 5√2

A(-1, -2) and D(-3, 0) are separated by a distance of [(-3 - (-1))²].

+ (0 - (-2))²]

= √[(-2)² + (2)²]

= √(4 + 4) = √8 AD = 2√2

Evidently, AB equals CD and BC = AD.

Both sides are equally valid.

As a result, a rectangle is formed using the specified points.

Explanation:

To identify the type of triangle, we must first establish the lengths of all three sides and check to see if they satisfy the triangle's requirements.

Using a calculation for the gap from both places,

AB=√(−4+5²2+(−2−6)²=√(1)2+(−8)²=√1+64=√65       [∵d=√(x₂−x₁)+(y₂−₁1)²]BC=√(7+4)2+(5+2)²=√(11)2+(7)²=√121+49=√170And CA=√(−5−7)²+(6−5)²=√(−12)²+(1)2√144+1=√145

We can observe that ABC does not meet the Pythagoras theorem.

Because each side is a distinct length, the necessary triangle is scalene.

Explanation:

When P = (x, 0)

Since A=(7, 4)

AP=25 (x−7)²+4²=20 (x−7)²=20−16=4

x−7−2x−7=−2 x=9x=5

(9,0)(5,0)

∴There are 2 equivalent details: (9, 0) & (5, 0).

Explanation:

The measurement of each side of the supplied quadrilateral ABCD will be determined first.

AB=(7−2)²+(3+2)²=25+25=52

BC=(11−7)²+(−1−3)²=16+16=42

CD=(6−11)²+(−6+1)²=25+25=52

DA=(6−2)²+(−6+2)²=16+16=42

Here, we can see that the quadrilateral's two opposite sides are equal. Therefore, ABCD must be a parallelogram.

Now, in order to identify what kind of parallelogram this is, we will find the relationship between the diagonals.

ACBD=(11−2)²+(−1+2)²=81+1=82=(6−7)²+(−6−3)²=1+81=82

Therefore, ABCD is a rectangle since the diagonals of the parallelogram ABCD are equal.

Explanation:

Provided, there are 9 units between positions A(-3, -14) and B(a, -5).

We must ascertain what an is worth.

[(x₂ - x₁)] is the distance between two places P (x₁, y₁) and Q (x₂, y₂).² + (y₂ - y₁)²]

The distance now is [(a - (-3))] between positions A (-3, -14) and B (a, -5)² + (-5 - (-14))²] = 9 √[(a + 3)² + (-5 + 14)²] = 9 

√[(a + 3)² + (9)²] = 9

(a + 3) on squaring both sides² + 81 = 81 (a + 3)² = 81 - 81 (a + 3)² = 0

Using square root, a = -3 and a + 3 = 0

As a result, a has a value of -3.

Explanation:

with the measurement calculation 

(x+5)²+(y−4)²=(x+1)²+(y−6)²x²+y²+10x+(−8y)+25+16

=x²+y²+2x+(−12y)+36+18x+4y+4

=02x+y+1=0 The first one is (1,3). 

They are going to be points.

Explanation:

On the x-axis is the point Q.

Therefore, it will have a coordinate of the form (x,0).

Taken as Q(x,y)=(x,0), these locations are equally spaced from A(5,2) and B(4,2).

We use the distance formula to determine the distance.

d=(x₂−x₁)²+(y₂−y₁)² ⟹AQ=(x−x₁)²+(y−y₁)²=(x−5)²+(0−2)²

BQ=(x-x₂) and2+(y−y₂)²=(x−4)²+(0−2)However, AQ=BQ.

∴(x−5)²+(0−2)²=(x−4)²+(0−2)² ⟹x=9/2=4.5 ∴ Q's coordinates are (4.5, 0).

Q is located on the perpendicular bisector of AB, thus AQ=BQ.

The ABQ is an isosceles triangle as a result.

Explanation:

Given that they are collinear, the points (5, 1) (-2, -3) and (8, 2m) are.

The value of m must be determined.

A triangle's area is 1/2[x1(y₂ - y₃) + x2(y₃ - y₁) + x3(y₁- y₂)] when its vertices are A (x₁, y₁), B (x₂, y₂), and C (x₃, y₃).

The triangle's area must be zero in order to determine whether the points are collinear.

In this instance, (x₁, Y₁) = (5, 1), (x₂, Y₂) = (-2, -3), and (x3, Y3) = (8, 2m).

Triangle's area is equal to half of [5(-3 - 2m) + -2(2m - 1) + 8(1 - (-3))] 0 -13 - 14m + 32 = 0 -14m + 19 = 0 14m = 19 m = 19/14 = 0 -15 - 10m - 4m + 2 + 8(4)

M thus has a value of 19/14.

Explanation:

Because P(3, 8) and Q(-10, y) are both equally distant from the point A(2, -4)

We must determine the value of y and the PQ distance.

P and Q are equally far apart from point A.

Thus, PA = QA.

[(x2 - x1)] is the distance between two places P (x₁, y₁) and Q (x₂, y₂).² + (y₂ - y₁)²]

P(3, 8) and A(2, -4)'s distance is equal to [(2 - 3)2 + (-4 - 8)].²]

= √[(-1)² + (-12)²]

= √(1 + 144)

= √145

Q(-10, y) and A(2, -4)'s length is equal to [(2 - (-10))2 + (y - (-4))]²]

= √[(12)² + (y + 4)²]

= √144 + (y + 4)²]

Now, √145 = √144 + (y + 4)²]

When each side are squared, 145 = 144 + (y + 4)²

Algebraic identity yields (a - b).² = a² + b² - 2ab

145 = 144 + y² + 8y + 16

145 - 144 - 16 = y² + 8y -15 = y² + 8y

So, y² + 8y + 15 = 0

After factoring, y² + 5 + 3 + 15 = 0 and y(y + 5) + 3(y + 5) = 0 (y + 3) (y + 5) = 0 were obtained.

Now, y + 3 = 0 y = - 3

Also, y = -5 and y + 5 = 0

Consequently, y has values of -3 and -5.

[(-10 - 3)] is the distance between P(3, 8) and Q(-10, -3).² + (-3 - 8)²]

= √[(-13)² + (-11)²] = √(169 + 121) = √290 units

[(-10 - 3)2 + (-5 - 8)2] is the formula for the distance between P(3, 8) and Q(-10, -5)

= √[(-13)² + (-13)²] = √(169 + 169) = √338 units

PQ is therefore separated by 290 and 338 units.

Explanation:

Because the triangle's vertices are (-8, 4), (-6, 6), and (-3, 9), we must determine its area.

A triangle's area is 1/2[x1(y₂ - y₃) + x2(y₃ - y₁) + x₃(y₁ - y2)] when its vertices are A (x₁, y₁), B (x₂, y₂), and C (x₃, y₃).

In this instance, (x₁, Y₁) = (-8, 4), (x₂, Y₂)

 = (-6, 6), and (x₃, Y₃) = (-3, 9)

The triangle's area is equal to 1/2 [-8(6 - 9) + -6(9 - 4) + -3(4 - 6)]

= 1/2[-8(-3) - 6(5) - 3(-2)]

= 1/2[24 - 30 + 6]

 = 1/2[-6 + 6]

 = 0

The triangle's area is therefore equal to 0 square units.

Explanation:

The line segment connecting (-4, -6) and (-1, 7) is provided.

The line portion's division ratio and the place of the division's coordinates must be determined.

Using the section formula

[(kx₂ + x₁)/(k + 1), (ky₂+ y₁)/(k + 1)] are the coordinates of the point P(x, y), which internally splits the line segment between the points A (x₁, y₁) and B (x₂, y₂) in the ratio k: 1.

In this instance, (x₁, Y₁) = (-4, -6), and (x2, Y2) = (-1, 7).

For these reasons, [(k(-1) + (-4))/(k + 1), (k(7) + (-6))/(k + 1)] = k:1

[(-k - 4)/(k + 1), (7k - 6)/(k + 1)] = k:1

On the x-axis, the point is located. i.e.,y = 0

So, 7k - 6/k + 1 = 0

7k - 6 = 0

7k = 6

 k = 6/7

As a result, 6:7 is the divisional ratio.

Explanation:

Provided that, the line segment connecting points A(1/2, 3/2) and B(2, -5), is divided by point P(3/4, 5/12).

The ratio by which point P divides the line segment AB must be determined.

Using the section formula

[(kx₂ + x₁)/(k + 1), (ky₂ + y₁)/(k + 1)] are the coordinates of the point P(x, y), which internally splits the line segment between the points A (x₁, y₁) and B (x₂, y₂) in the ratio k: 1.

Here, (x₁, Y₁) = (1/2, 3/2) and (x₂, Y₂) = (2, -5) respectively.

[(k(2) + (1/2))/(k + 1) , (k(-5) + (3/2))/(k + 1)] = (3/4, 5/12)

[(2k + (1/2))/(k + 1), (-5k + (3/2))/(k + 1)] = (3/4, 5/12)

Therefore, (2k + (1/2))/(k + 1) = 3/4 (2k + (1/2))/(k + 1)

(4,000 + 1)/2 = (4,000 + 3)/4 (4,000 + 1) = (4,000 + 3)/2

2(4k + 1) = (4,000 + 3) + 2 = (4,000 + 3) + 8k- 3k = 3 - 2 

5k = 1

 k = 1/5.

Explanation:

Given, the line segments connecting A(3a + 1, -3) and B(8a, 5) are divided in the ratio 3:1 by point P(9a - 2, -b).

The values of a and b must be determined.

[(m₁x₂ + m₂x1)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)] are the coordinates of the point P that internally splits the line segment joining the points A (x₁, y₁) and B (x₂, y₂) in the ratio m₁: m₂.

In this case, m₁:m₂ = 3:1, (x₁, Y₁) = (3a + 1, -3), and (x₂, Y₂) = (8a, 5)

So, [(3(8a) + 1(3a + 1))/(3 + 1), (3(5) + 1(-3))/(3 + 1)] = (9a - 2, -b)

[(24a + 3a + 1)/4, (15 - 3)/4] = (9a - 2, -b)

[(27a + 1)/4, 12/4] = (9a - 2, -b)

Consequently, 27a + 1 = 4(9a - 2) (27a + 1)

27a + 1 = 36a - 8 36a - 27a = 1 + 8 9a= 9

 a = 9/9

 a = 1

Additionally, b = -3

b = -3

As a result, a and b have values of 1 and -3.

Explanation:

Given that, the line segment between points A(10, -6) and B(k, 4) has a midpoint at (a, b).

Additionally, a - 2b = 18 ----------- (1)

Finding k's value and AB's distance is required.

[(x₁ + x₂)/2, (y₁ + y₂)/2] are the coordinates of the midpoint of the line segment connecting the points P (x₁, y₁) and Q (x₂, y₂).

In this instance, (x₁, Y₁) = (10, -6), and (x₂, Y₂) = (k, 4).

[(10 + k)/2, (-6 + 4)/2] is the midpoint of AB. = (a, b) [(10 + k)/2, -2/2] = (a, b)

At this point, (10 + k)/2 = a 10 + k = 2 a - 10 ---------------- (2)

Furthermore, b = -2/2 b = -1

Add b = -1 to (1).

a - 2(-1) = 18 a + 2 = 18 a = 18 - 2 a = 16

Add a = 16 to (2) k.= 2(16) - 10

 k = 32 - 10

 k = 22

K thus has a value of 22.

the separation between two points P (x₁, y₁) and Q (x₂, y₂) are expressed as [(x₂ - x₁)² + (y₂ - y₁)²]

Distance between points A(10, -6) and B(22, -4) is equal to [(22 - 10)2 + (4 - (-6))2].

= √[(12)² + (10)²]

= √(144 + 100) = √244 = 2√61

As a result, A and B are separated by 261 units.

Explanation: 

Let's say that the provided location P(x₁,y₁)=(11,9) lies on a circle with a radius of r and a center at C(x₂,y₂)=(2a,a7). assuming the circle's diameter is 102 units, PC=r. (i) Then, r=2 diameter=2102 units=52 units. (ii). Once more, the length calculation is d=(x₁ x₂)2+(y₁ y₂)²; PC=d=(112a)²+(9a+7)²=5a240a+125

From (i)&(ii), we have 5a240a+125=(52)², 

5a240a+75=0, (a5)(5a3)=0, a=5,53, and Ansa=5, ⅗.

Explanation:

Provided that there is a 1:2 division at the location P on the line segment connecting points A(3, 2) and B(5, 1)

On line 3x - 18y + k = 0, point P is located.

The value of k must be determined.

[(m₁x₂+m₂x₁)/(m₁+m₂), (m₁y2+m2y₁)/(m₁+m2)] are the values of the point P that internally splits the line segment joining the points A (x₁, y₁) and B (x₂, y₂) in the ratio m₁: m₂.

In this instance, m₁:m₂ = 1:2, (x₁, Y₁) = (3, 2), and (x₂, Y₂) = (5, 1).

Let P be the point (x, y) [(1(5) + 2(3)), (1(1) + 2(2))/(1 + 2)] = (x, y)

[(5 + 6)/3, (1 + 4)/3] = (x, y)

 [11/3, 5/3] = (x, y)

P is located at (11/3, 3/3).

Point P is located on the line 3(11/3) - 18(5/3) + k = 0

Consequently, 11 - 6(5) + k = 0

11 - 30 + k = 0 

k = 19.

Explanation:

The area of the triangle created by linking a triangle's midpoints is 14.

DEF area equals 14 ABC area

Area of DEF = 4 times Area of ABC

Triangle with sides (x₁, y₁), (x₂, y₂), and (x₃, y₃) has an area.

Area is equal to 12[(x₁ y₂-x₂y₁)+(x₂y₃-x₃y₂)+(x₃y₁-x₁y₃)].

DEF area equals 12[⅗+,35/2+49/2+ 21/2+35/44+7/10]

DEF's area is 12[10720] = 10740 square units.

ABC's area is 4 (DEF's area is 4 x 10740), which is 10.7 square units.

Explanation:

The vertices of a triangle ABC with a right angle at B are A (2, 9), B(a), and C (5, 5), respectively.

We must determine the area of the triangle ABC and the value of a.

Because B is at a right angle in the triangle ABC, it is a right triangle.

According to Pythagoras' Theorem, AC2 = AB2 + BC2.

[(x₂ - x₁)² + (y₂ - y₁)²] is the distance between two places P (x₁, y₁) and Q (x₂, y₂).

The separation of A(2, 9) and C(5, 5) = √[(5 - 2)² + (5 - 9)²]

= √[(3)² + (-4)²]

= √(9 + 16) = √25 = 5

A(2, 9) and B(a, 5) are separated by [(a - 2)].² + (5 - 9)²]

= √[(a - 2)² + (-4)²]

= √[(a - 2)² + 16]

[(5 - a)2 + (5 - 5)2] is the formula for the distance between B(a, 5) and C(5, 5).

= √[(5 - a)² + 0]

= √(5 - a)²

Now, (5)² = (√[(a- 2)² + 16])² + (√(5 - a)²)²

25 = (a - 2)² + 16 + (5 - a)²

Algebraic identity demonstrates that (a - b)² = a² - 2ab + b².

25 = a² - 4a + 4 + 16 + 25 - 10a + a2 at this point.

25 = 2a² -14a + 45

2a² - 14a + 45 - 25 = 0

2a² - 14a + 20 = 0

By dividing by 2, a2 - 7 a + 10 equals 0.

A2 - 5a - 2a + 10 = 0 on factoring a(a - 5) - 2(a - 5) = 0 (a - 2)(a - 5) = 0

Now, a - 2 = 0 a = 2

Also, a = 5 and a - 5 = 0.

A has values of 2 and 5.

Distance BC = ((5 - 5)2 = 0 when a = 5)

A = 5 is therefore not conceivable.

Consequently, a has a value of 2.

1/2 [x1(y2] is the area of a triangle with the vertices A (x₁, y₁), B (x₂, y₂), and C (x₃, y₃).

Explanation:

The given location R is located on the line segment connecting P(-1, 3), and Q(2, 5).

The location of point R must be determined so that PR = 3/5 PQ.

PR/PQ = 3/5

PQ/PR = 5/3 on rearrangement

PQ = PR + QR, as we are aware.

This means that (PR + QR) / PR = 5/3 

PR/PR + QR/PR = 5/3 

1 + QR/PR = 5/3

 QR/PR = 5/3 - 1 QR/PR = 5-3/3 QR/PR = 2/3

Rearranging once more, PR/QR = 3/2 PR: QR = 3:

As a result, the line segment PQ is divided by the point R in a 3:2 ratio.

[(m₁x₂ + m₂x₁)/m₁ + m₂, (m₁y₂ + m₂y₁)/m₁ + m₂] are the coordinates of the point P that internally splits the line segment joining the locations A (x₁, y₁) and B (x₂, y₂) in the ratio m₁: m₂.

As shown, m₁: m₂ = 3: 2.(x₁, Y₁) = (-1, 3), and (x₂, Y₂) = (-2, 5)

R's coordinate should be (x, y).

(x, y) = [(3(2) + 2(-1))/3 + 2, (3(5) + 2(3))/3 + 2]

(x, y) = [(6 - 2)/5, (15 + 6)/5]

(x, y) = [4/5, 21/5]

As a result, point R's coordinates are (4, 21).

Explanation: 

Because they are collinear, the points A(k + 1, 2k), B(3k, 2k + 3), and C(5k - 1, 5k) are congruent.

We must ascertain the values of k.

The triangle's area must be zero in order to determine whether the points are collinear.

A triangle's area is 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)] when its vertices are A (x₁, y₁), B (x₂, y₂), and C (x₃, ₃3).

In this case, (x₁, Y₁) equals (k + 1, 2k), (x₂, Y₂) equals (3k, 2k + 3), and (x3, Y3) equals (5k - 1, 5k).

The triangle's area is equal to 1/2 [(k + 1)(2k + 3 - 5k) + 3k(5k - 2k) + (5k - 1)(2k - (2k + 3))] = 0

1/2[(k + 1)(3 - 3k) + 3k(3k) + (5k - 1)(-3)] = 0

According to the distributive and multiplicative properties, 3k - 3k2 + 3 - 3k + 9k2 - 15k + 3 = 0 9k2 by grouping.

Explanation:

The right answer is A 8: 1.

Let k = 1 be the ratio.

An (8,-9) and B(2,1) should be divided in the ratio k:1 by P (x, y).

By using the equations P (x, y) =2k+8/k+1, k-9/k+1,

Additionally, P(2k+8/k+1, k-9+/k+1) lies on the line 2x+3y5=0.

∴ 2(2k+8)/k+1)+3(k-9)/k+1−5=0

4k+16/k+1 +3k-27/k+1−5=0

4k+16+3k−27−5k−5=0

⇒2k=16 ⇒k=8

Line 2x+3y5=0 splits the line section between points A(8, -9) and B(2, 1) in an 8:1 ratio.

Explanation:

If triangle A(4,3) and B(4,3) are provided

The triangle is parallel if C(x,y) 

AC=BC (x+4)²+(y−3)²=(x−4)²+(y−3)²

(x+4)²+(y−3)²=(x−4)²+(y−3)²

x2+16+8x=x2+16−8x

x=0

So, using the measurement calculation, C(0,y) AB=8

AC=8 (4)²+(y−3)²=8

16+y²+9−6y=64

y²−6y−39=0

y=26±36+156

y=26±192

y=3±43

For the origin, point C(0,3-43) is the interior. and C(0,3+43) for exterior origin.

Explanation: 

ABCD is a parallelogram, so there.

The parallelogram's vertices are A(6, 1), B(8, 2), and C(9, 4).

E is where DC is centered.

We must calculate the triangle's area, ADE.

Let (x, y) serve as the parallelogram's fourth vertex.

A parallelogram's diagonals are known to be split in half.

Midpoint of AC = Midpoint of BD, for example.

[(x₁ + x₂)/2, (y₁ + y₂)/2] are the coordinates of the midpoint of the line segment connecting the points P (x₁, y₁) and Q (x₂, y₂).

A(6, 1) and C(9, 4)'s midpoint is [(6 + 9)/2, (1 + 4)/2] = [15/2, 5/2].

B(8, 2) and D(x, y)'s midpoint is [(8 + x)/2, (2 + y)/2].

Now, [(8 + x)/2, (2 + y)/2] = [15/2, 5/2]

(8 + x)/2 = 15/2 8 + x = 15

x = 15 - 8 x = 7 (2 + y)/2 = 5/2

2 + y = 5 y = 5 - 2 y = 3

Consequently, the fourth vertex's coordinates are D = (7, 3).

D(7, 3) and C(9, 4) midpoint equals E.

E = [(7 + 9)/2, (3 + 4)/2]

E = [16/2, 7/2]

E = [8, 7/2]

A triangle's area is 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x3(y₁ - y₂)] when its vertices are A (x₁, y₁), B (x₂, y₂), and C (x₃, y₃).

In this case, (x₁, Y₁) = (6, 1), (x₂, Y₂) = (7, 3), and (x₃, Y₃) = (8, 7/2) = (8, 3.5).

Triangle ADE's area is equal to half of [6(3-3.5) + 7(3.5-1) + 8(1-3)]

= 1/2[6(-0.5) + 7(2.5) + 8(-2)]

= 1/2[-3 + 17.5 - 16]

= 1/2[-19 + 17.5] = 1/2[-1.5] = -0.75

There can be no negative area.

So, the area is equal to 0.75 square units.

Consequently, the region of.

Explanation:

Given, A (x₁, y₁), B (x₂, y₂), and C (x₃, y₃) are the vertices of ABC.

When BQ:QE = 2:1 and CR:RF = 2:1, respectively, the points Q and R are the points on the medians BE and CF.

The coordinates of the points Q and R must be determined.

Let Q's coordinates be (a, b) and R's coordinates be (p, q).

[(x₁ + x₂)/2, (y₁ + y₂)/2] are the coordinates of the midpoint of the line segment connecting the points P (x₁, y₁) and Q (x₂, y₂).

The midpoint between A(x₁, ₁) and C(x3, y3) is E.

E = [(x₁ + x₃)/2, (y₁ + y₃)/2]

The midpoint between A(x₁, y₁) and B(x, y) is F.

F = [(x₁ + x₂)/2, (y₁ + y₂)/2]

Using the section formula

[(m₁x₂+ m₂x₁)/m₁ + m₂, (m₁y₂ + m₂y₁)/m₁ + m₂] are the coordinates of the point P that internally splits the line segment joining the locations A (x₁, y₁) and B (x₂, y₂) in the ratio m₁: m2.

The lines B(x₂, y₂) and E((x₁ + x3)/2, (y₁ + y3)/2) are divided by the point Q in the ratio 2:1.

Q's coordinates are [2(x₁ + x3/2 + 1(x₂)/2 + 1, 2(y₁+ y₃)/2 + 1(y₂)/2 + 1]

(a, b) = [(x₁, x₂, and x₃)/3, (y₁, y₂, and y₃)/3]

The line C(x₃, y₃) and F((x₁ + x₂)/2, (y₁ + y₂)/2) are divided by the point R in the ratio 2:1.

R, (p), and q coordinates are [2(x₁ + x₃)/2 + 1(x₂)/2 + 1, 2(y1 + y₃)/2 + 1(y₂)/2 + 1]

(p, q) is equal to [(x₁, x2, x₃)/3, (y₁, y₂, y₃)/3].

As a result, Q and R's coordinates are ((x₁ + x2 + x₃)/3, (y1 + y₂ + y₃)/3) and ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3) respectively.

Explanation: 

Provided, a parallelogram is formed by the points A(1, -2) B(2, 3), C(a, 2), and D(-4, -3).

Using AB as the foundation, we must determine the amount of a and the height of the parallelogram.

A parallelogram's diagonals are known to be split in half.

As a result, the parallelogram ABCD's diagonals AC and BD are split in half.

Thus, the midway of AC equals the midpoint of BD.

[(x₁ + x₂)/2, (y₁ + y₂)/2] are the coordinates of the midpoint of the line segment connecting the points P (x₁, y₁) and Q (x₂, y₂).

A(1, -2) and C(a, 2)'s midpoint is [(1 + a)/2, (-2 + 2)/2] = [(1 + a)/2, 0]

B(2, 3) and D(-4, -3)'s midpoints are equal to [(2 - 4)/2, (3 - 3)/2] 

= [(-2/2), 0] 

= [-1, 0].

Now, [(1 + a)/2, 0] = (-1, 0)

Consequently, 1 + a/2 = -1 1 + a = -2

a = -2 - 1 

a = -3

The volume of the parallelogram ABCD now equals 2 (the area of the triangle ABC).

A triangle's area whose vertices are (x1, y1)

B (x₂, y₂) and C (x3, y3) are 1/2 [x1(y₂, y3) + x₂(y3, y1) + x3(y₂, y₂)].

In this case, (x1, y1) = (1, -2) As well as (x₂, y₂) = (2, 3), (x3, y3) = (-3, 2)

Triangle ABC's area is now 1/2[1(3 - 2) + 2(2 - (-2) + -3(-2 - 3)]

= 1/2[(1) + 2(4) - 3(-5)]

= 1/2[1 + 8 + 15]

1/2[24] equals 12 square units.

Parallelogram's area is equal to 2(12) or 24 square units.

AB = Base

[(x₂ - x1)2 + (y₂ - y1)2] is the distance between two places P (x1, y1) and Q (x₂, y₂).

Distance [(2 - 1)2 + (3 - (-2))2] between points A(1, -2) and B(2, 3)

= √[(1)² + (5)²]

= √(1 + 25) = √26

We are aware that the parallelogram's area is equal to its base plus its height.

24 equals 26 inches tall

Dimensions = 24/26

The parallelogram's height is 24/26 units as a result.

Explanation: 

assuming that ABCD is a square with A(x1,y1)=(3,5), B(x₂,y₂)=(7,8), C(x3,y3)=(11,5), and D(x4,y4)=(7,2), the correct choice is A.∴Jaspal will be situated at themed point M(x,y) between ACorBD.A square is (AC=BDABCD).By using either M(x,y)=(2x1+x3,2y1+y3) or M(x,y)=(x1+x3/2, y1+y3/2) as the section formula for the themed location.⟹M(x,y)=(x2+x4/2, y2+y4/2)orM(3+11/2, 5+5/2)⟹M7+7/2, 8+2/2.

The coordinates of Jaspalis are therefore (7, 5). The coordinates of Jaspalis are (7, 5).

Explanation: 

By following the guidelines, we created a diagram in which each location is marked with its coordinates and direction.

We are aware that d=(x₂x₁) is the amount of time between two places (x₁,y₁) and (x₂,y₂)²+(y₂−y₁)²

The gap between the house and the bank is now (52)²+(8−4)² =√(3)²+(4)²=√9+16=√25=5

Bank is (135) miles away from the daughter's school.2+(14−8)2=√(8)2+(6)2=√64+36=√100=10

The length of the road to my office and my daughter's school is (13 + 13)²+(26−14)²=√0+(12)²=12

Houses, banks, Schools, and Offices travel a total distance of 5 units, 10 units, 12 units, and 27 units.

Distance from home to offices is equal to (132)²+(26−4)²=√(11)²+(22)²=√121+484=√605=24.59≈24.6 km

Ayush had to travel an extra 2.4 kilometers to get to his office, which is equal to 27 minus 24.6.

Ayush will therefore need to go an additional 2.4 kilometers.