Equations of Straight Lines: practice questions

6 questions · 12 marks · free, no account · Read the notes first

1. Question 1Multiple choice · 1 mark

   Find the slope of the straight line $3x - 4y + 12 = 0$.

   A

   $-\dfrac{3}{4}$

   B

   $\dfrac{4}{3}$

   C

   $\dfrac{3}{4}$

   D

   $-\dfrac{4}{3}$

2. Question 2Multiple choice · 1 mark

   Find the equation of the straight line passing through $(2, 3)$ and perpendicular to $y = 2x + 1$.

   A

   $x + 2y - 8 = 0$

   B

   $2x - y - 1 = 0$

   C

   $x - 2y + 4 = 0$

   D

   $x + 2y - 4 = 0$

3. Question 3Multiple choice · 1 mark

   Find the $x$-intercept of the straight line $2x + 5y - 10 = 0$.

   A

   $2$

   B

   $-5$

   C

   $10$

   D

   $5$

4. Question 4Multiple choice · 1 mark

   $A(1, 2)$ and $B(7, -4)$. $P$ divides $AB$ internally in the ratio $AP : PB = 1 : 2$. Find the coordinates of $P$.

   A

   $(5, -2)$

   B

   $(3, 0)$

   C

   $(4, -1)$

   D

   $(3, -2)$

5. Question 54 marks

   The straight line $L_1$ is $2x - y + 3 = 0$. The straight line $L_2$ passes through $(4, 1)$ and is perpendicular to $L_1$.

   (a) Find the equation of $L_2$.

   (b) Find the coordinates of the point of intersection of $L_1$ and $L_2$.

   Show worked solution

   (a) The slope of $L_1$ is $2$, so the slope of $L_2$ is $-\dfrac{1}{2}$. 1M

   $y - 1 = -\dfrac{1}{2}(x - 4)$, i.e. $x + 2y - 6 = 0$ 1A

   (b) Substituting $y = 2x + 3$ into $L_2$: $x + 2(2x + 3) - 6 = 0$ 1M

   The point of intersection is $(0, 3)$. 1A

6. Question 64 marks

   $A(-2, 1)$ and $B(4, 5)$. Find the equation of the perpendicular bisector of $AB$.

   Show worked solution

   Mid-point of $AB = (1, 3)$ 1A

   Slope of $AB = \dfrac{2}{3}$, so the slope of the perpendicular bisector is $-\dfrac{3}{2}$. 1M

   $y - 3 = -\dfrac{3}{2}(x - 1)$ 1M

   $3x + 2y - 9 = 0$ 1A