Answer:
[tex]\textsf{a)} \quad 3x+2y-19=0[/tex]
b) (i) N = (1, 8)
b) (ii) ΔMNQ = 9 units²
Step-by-step explanation:
Given points:
- P = (-5, 8)
- Q = (7, 8)
- R = (-1, 2)
Part (a)
If M is the midpoint of QR then:
[tex]\implies \textsf{M}=\left(\dfrac{x_R+x_Q}{2},\dfrac{y_R+y_Q}{2}\right)[/tex]
[tex]\implies \textsf{M}=\left(\dfrac{-1+7}{2},\dfrac{2+8}{2}\right)[/tex]
[tex]\implies \textsf{M}=\left(3,5\right)[/tex]
If line l is parallel to PR, then its gradient is the same as the gradient of line PR. Therefore, find the gradient of line PR:
[tex]\implies \textsf{gradient}\:(m)=\dfrac{y_R-y_P}{x_R-x_P}=\dfrac{2-8}{-1-(-5)}=-\dfrac{3}{2}[/tex]
Substitute the found gradient and the coordinates of point M into the point-gradient formula:
[tex]\implies y-y_M=m(x-x_M)[/tex]
[tex]\implies y-5=-\dfrac{3}{2}(x-3)[/tex]
[tex]\implies 2y-10=-3x+9[/tex]
[tex]\implies 3x+2y-19=0[/tex]
Part (b)
(i) Point N is the point of intersection of line l and PQ.
- line l: 3x+2y-19=0
- line PQ: y = 8
To find the x coordinate of point N, substitute y = 8 into the equation for line l and solve for x:
[tex]\implies 3x+2(8)-19=0[/tex]
[tex]\implies 3x-3=0[/tex]
[tex]\implies 3x=3[/tex]
[tex]\implies x=1[/tex]
Therefore, the coordinates of point N are (1, 8).
(ii)
Triangle MNQ
[tex]\textsf{Height} = y_N-y_M=8-5=3[/tex]
[tex]\textsf{Base} = x_Q-x_N=7-1=6[/tex]
[tex]\begin{aligned}\textsf{Area of a triangle} & = \sf \dfrac{1}{2} \times base \times height\\\\\implies \textsf{Area of triangle MNQ} & = \sf \dfrac{1}{2} \times 6 \times 3\\& = \sf 9\:\:units^2 \end{aligned}[/tex]
