Question

which machines will you use to do the following work ? 1. to load a heavy drum onto the truck​

Answer 1

Answer:

$\large\underline{\sf{Solution-}}$

Given that,

↝ Triangle ABC is an isosceles triangle such that AB = AC.

↝ Coordinates of A is (4,3) and B is (1,-1)

↝ Gradient of the straight line BC is 1/2.

↝ AD is perpendicular to BC.

Part (a) Equation of BC

As,

Line BC passes through the point (1, - 1) and having gradient or slope 1/2.

We know,

Equation of line which passes through the point (a, b) and having gradient or Slope m is

$\rm :\longmapsto\:\boxed{ \tt{ \: \: y \: - \: b \: = \: m \: (x \: - \: a) \: \: }}$

So, using this,

The equation of BC is

$\rm :\longmapsto\:y - ( - 1) = \dfrac{1}{2}(x - 1)$

$\rm :\longmapsto\:y + 1 = \dfrac{1}{2}(x - 1)$

$\rm :\longmapsto\:2y + 2 = x - 1$

$\red{\rm \implies\:\boxed{ \tt{ \: x - 2y = 3 \: }}} - - - - (1)$

Part (a) Equation of AD

Now, it is given that AD is perpendicular to BC.

We know,

Two lines having slope m and M are perpendicular iff Mm = - 1.

So, given that,

↝ Slope or gradient of BC = 1/2.

↝ So, gradient of AD = - 2.

Now, Line AD passes through the point (4, 3) and having gradient - 2.

So, Equation of AD is

$\rm :\longmapsto\:y - 3 = - 2(x - 4)$

$\rm :\longmapsto\:y - 3 = - 2x + 8$

$\red{\rm \implies\boxed{ \tt{ \: \:2x + y = 11}}} - - - - - (2)$

Part (b) Coordinates of D

Now, D is the point of intersection of line AD and BC.

So, Solve the system of linear equations.

From equation (1), we have

$\rm :\longmapsto\:x - 2y = 3$

$\rm :\longmapsto\:x = 2y + 3$

Substituting this value in equation (2), we get

$\rm :\longmapsto\:2(2y + 3) + y = 11$

$\rm :\longmapsto\:4y + 6+ y = 11$

$\rm :\longmapsto\:5y = 5$

$\bf\implies \:y = 1$

On substituting y = 1 in equation (1), we get

$\rm :\longmapsto\:x - 2 = 3$

$\rm :\longmapsto\:x = 3 + 2$

$\bf\implies \:x = 5$

Hence, Coordinates of D is (5, 1).

Part (b) Coordinates of C

As we have with us

↝ Coordinates of B (1, - 1)

↝ Coordinates of D (5, 1)

↝ Also, AB = AC

↝ So, AD is perpendicular bisector of BC.

It means, D is the midpoint of BC.

Let assume that coordinates of C be (a, b).

So, using Midpoint Formula, we have

$\rm :\longmapsto\:(5,1) = \bigg(\dfrac{a + 1}{2}, \: \dfrac{b - 1}{2} \bigg)$

So, on comparing,

$\rm :\longmapsto\:\dfrac{a + 1}{2} = 5 \: \: and \: \: \dfrac{b - 1}{2} = 1$

$\rm :\longmapsto\:a + 1 = 10 \: \: and \: \: b - 1 = 2$

$\rm :\longmapsto\:a = 10 - 1\: \: and \: \: b = 2 + 1$

$\bf\implies \:a = 9 \: \: and \: \: b = 3$

Hence, Coordinates of C is (9, 3).