Question

6.

P(-5, 8)
R(-1,2)
0
The line / cuts PQ at the point N.
(b) Find
M
Figure 3
The points P(-5, 8), Q(7, 8) and R(-1, 2) form the vertices of a triangle PQR, as shown
in Figure 3. The point M is the midpoint of QR.
The line / passes through M and is parallel to PR.
(a) Find an equation for 1, writing your answer in the form ax + by + c = 0,
where a, b and c are integers to be found.
(4)
(i) the coordinates of N,
(ii) the area of triangle MNQ.
Q(7,8)
(3)
Leave
blank

Answer 1

To find the equation of the line / that passes through the midpoint of QR and is parallel to PR, we first find the midpoint of QR. The x-coordinate of the midpoint is 3 and the y-coordinate of the midpoint is 5. The equation of the line / is 3x - 4y + 11 = 0.

Step-by-step explanation:

To find the equation of the line / that passes through the midpoint of QR and is parallel to PR, we first find the midpoint of QR. The x-coordinate of the midpoint is (7+(-1))/2 = 6/2 = 3, and the y-coordinate of the midpoint is (8+2)/2 = 10/2 = 5. So the midpoint of QR is M(3, 5).

Since the line / is parallel to PR, it will have the same slope as PR. The slope of PR can be found using the formula (y2-y1)/(x2-x1), where (x1, y1) = (-1, 2) and (x2, y2) = (7, 8). So the slope of PR is (8-2)/(7-(-1)) = 6/8 = 3/4.

Now we can use the point-slope form of a line to find the equation of the line /. The equation is y - y1 = m(x - x1), where (x1, y1) = (3, 5) and m = 3/4. Substituting the values, we have y - 5 = (3/4)(x - 3). Multiplying through by 4, we get 4y - 20 = 3x - 9. Rearranging the terms, the equation of the line / is 3x - 4y + 11 = 0.

Answer 2

`\textsf{a)} \quad 3x+2y-19=0`

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:

`\implies \textsf{M}=\left((x_R+x_Q)/(2),(y_R+y_Q)/(2)\right)`

`\implies \textsf{M}=\left((-1+7)/(2),(2+8)/(2)\right)`

`\implies \textsf{M}=\left(3,5\right)`

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:

`\implies \textsf{gradient}\:(m)=(y_R-y_P)/(x_R-x_P)=(2-8)/(-1-(-5))=-(3)/(2)`

Substitute the found gradient and the coordinates of point M into the point-gradient formula:

`\implies y-y_M=m(x-x_M)`

`\implies y-5=-(3)/(2)(x-3)`

`\implies 2y-10=-3x+9`

`\implies 3x+2y-19=0`

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:

`\implies 3x+2(8)-19=0`

`\implies 3x-3=0`

`\implies 3x=3`

`\implies x=1`

Therefore, the coordinates of point N are (1, 8).

(ii)

Triangle MNQ

`\textsf{Height} = y_N-y_M=8-5=3`

`\textsf{Base} = x_Q-x_N=7-1=6`

`\begin{aligned}\textsf{Area of a triangle} & = \sf (1)/(2) * base * height\\\\\implies \textsf{Area of triangle MNQ} & = \sf (1)/(2) * 6 * 3\\& = \sf 9\:\:units^2 \end{aligned}`