Question

Triangle PQR where P(1,4), Q(-3,-4),R(7,k) is right-angled at Q. find the value of k

Answer 1

Answer:

k = -9.

Step-by-step explanation:

As the triangle is right-angled at Q, by Pythagoras:

PR^2 = PQ^2 + RQ^2

So, substituting the given data and using the distance formula between 2 points:

(7 - 1)^2 + (k - 4)^2 =  (-4-4)^2 + (-3-1)^2 + (7 - (-3))^2 + (k - (-4))^2

36 + (k - 4)^2 = 64 + 16 + 100 + ( k + 4)^2

(k - 4)^2 - (k + 4)^2 = 180 - 36

k^2 - 8k + 16 - (k^2 + 8k + 16) = 144

-16k = 144

k = -9.

Answer 2

Answer:

k = - 9

Step-by-step explanation:

Since the triangle is right- angled at Q then PQ is perpendicular to RQ

Calculate the slope of PQ using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = P(1, 4) and (x₂, y₂ ) = Q(- 3, - 4)

$m_{PQ}$ = $\frac{-4-4}{-3-1}$ = $\frac{-8}{-4}$ = 2

Given a line with slope m then the slope of a line perpendicular to it is

$m_{perpendicular}$ = - $\frac{1}{m}$ = - $\frac{1}{2}$ , thus

$m_{RQ}$ = - $\frac{1}{2}$

Calculate the slope of RQ and equate to - $\frac{1}{2}$

$m_{RQ}$ = $\frac{k+4}{7+3}$ = - $\frac{1}{2}$ , that is

$\frac{k+4}{10}$ = - $\frac{1}{2}$ ( multiply both sides by 10 )

k + 4 = - 5 ( subtract 4 from both sides )

k = - 9