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:

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

Answer 2

The right-angled triangle PQR has two perpendicular sides PQ and QR, such that the value of k is -9

Given that:

$P =(1,4)$

$Q = (-3,-4)$

$R = (7,k)$

Calculate the slope of PQ using:

$m_1 = \frac{y_2 - y_1}{x_2 - x_1}$

$m_1 = \frac{-4-4}{-3-1}$

$m_1 = \frac{-8}{-4}$

$m_1=2$

Calculate the slope of QR using:

$m_2 = \frac{y_2 - y_1}{x_2 - x_1}$

$m_2 = \frac{k--4}{7--3}$

$m_2 = \frac{k+4}{7+3}$

$m_2 = \frac{k+4}{10}$

Because the triangle is right-angled at Q, then sides PQ and QR are perpendicular.

The relationship between perpendicular sides is:

$m_1 \times m_2 = -1$

So, we have:

$2 \times (\frac{k + 4}{10}) = -1$

$\frac{k + 4}{5} = -1$

Multiply through by 5

$k +4 = -5$

Subtract 4 from both sides

$k = -9$

Hence, the value of k is -9

Read more about right-angled triangles at:

https://brainly.com/question/3770177