Question

P is the point (a, a -2) and Q is the point (4-3a, -a).

a
Find the gradient of the line PQ.
b
Find the gradient of a line perpendicular to PQ.
Given that the distance PQ is 10√5, find the two possible values of a.
c

Answer 1

`\textsf{a)} \quad (1)/(2)`

`\textsf{b)} \quad -2`

`\textsf{c)} \quad a = -4,\;\; a = 6`

Explanation:

Part (a)

`\boxed{\begin{minipage}{9cm}\underline{Gradient Formula}\\\\Gradient $=(y_2-y_1)/(x_2-x_1)$\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are two points on the line.\\\end{minipage}}`

Given points:

• P = (a, a-2)
• Q = (4-3a, -a)

Substitute the given points into the gradient formula to find the gradient of line PQ:

`\textsf{Gradient}=(-a-(a-2))/((4-3a)-a)=(-a-a+2)/(4-3a-a)=(2-2a)/(4-4a)=(2(1-a))/(4(1-a))=(2)/(4)=(1)/(2)`

Part (b)

If two lines are perpendicular to each other, their gradients are negative reciprocals.

Therefore, the gradient of a line perpendicular to PQ is:

`\implies \textsf{Gradient}=-2`

Part (c)

`\boxed{\begin{minipage}{7.4 cm}\underline{Distance between two points}\\\\$d=√((x_2-x_1)^2+(y_2-y_1)^2)$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}`

Substitute the points and the given distance 10√5 into the formula and solve for a.

`\implies 10√(5)=√(((4-3a)-a)^2+(-a-(a-2))^2)`

`\implies 10√(5)=√((4-3a-a)^2+(-a-a+2)^2)`

`\implies 10√(5)=√((4-4a)^2+(2-2a)^2)`

`\implies 500=16-32a+16a^2+4-8a+4a^2`

`\implies 500=20a^2-40a+20`

`\implies 25=a^2-2a+1`

`\implies a^2-2a-24=0`

`\implies a^2-6a+4a-24=0`

`\implies a(a-6)+4(a-6)=0`

`\implies (a+4)(a-6)=0`

Apply the zero-product property:

`\implies a+4=0 \implies a=-4`

`\implies a-6=0 \implies a=6`

Therefore, the two possible values of a are:

• a = -4, a = 6