Straight-Line Graphs (Cambridge International IGCSE Further Maths): Exam Questions

Variables $x$ and $y$ are such that, when fourth root of y is plotted against $\frac{1}{x}$, a straight line graph passing through the points (0.5, 9) and (3, 34) is obtained. Find $y$ as a function of $x$.

Solutions to this question by accurate drawing will not be accepted.

Find the equation of the perpendicular bisector of the line joining the points $(4,-7)\mathrm{and}(-8,9)$.

Solutions by accurate drawing will not be accepted. The points $A\mathrm{and}B$ have coordinates (-2, 4) and (6,10) respectively.

Find the equation of the perpendicular bisector of the line $AB$, giving your answer in the form $ax+by+c=0$, where $a,b\mathrm{and}c$ are integers.

Extract:

The points A and B have coordinates (-2, 4) and (6, 10) respectively.

Find the equation of the perpendicular bisector of the line AB, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

The equation of the perpendicular bisector is 4x + 3y - 29 = 0

The point $C$ has coordinates $(5,p)$ and lies on the perpendicular bisector of $AB$. Find the value of $p$.

Extract:

The points A and B have coordinates (-2, 4) and (6, 10) respectively.

Midpoint of AB = (2, 7)

The point C has coordinates (5, p) and lies on the perpendicular bisector of AB.

From the equation 4x + 3y - 29 = 0, when x = 5: (4 × 5) + 3y - 29 = 0 20 + 3y - 29 = 0 3y = 9 y = 3

Therefore C = (5, 3)

It is given that the line $AB$ bisects the line $CD$. Find the coordinates of $D$.

Find the equation of the perpendicular bisector of the line joining the points (12, 1) and (4, 3), giving your answer in the form $y=mx+c$.

Extract:

Find the equation of the perpendicular bisector of the line joining the points (12, 1) and (4, 3), giving your answer in the form y space equals space m x plus c.

The equation of the perpendicular bisector is y equals 4 x minus 30

The perpendicular bisector cuts the axes at points $A$ and $B$. Find the length of $AB$.

Solutions to this question by accurate drawing will not be accepted. The points $A$ and $B$ are (4, 3) and (12, −7) respectively.

Find the equation of the line $L$, the perpendicular bisector of the line $AB$.

Extract:

The points A and B are (4, 3) and (12, −7) respectively.

Find the equation of the line L, the perpendicular bisector of the line AB.

From the mark scheme: Calculate the gradient between points A and B: gradient_AB = (-7 - 3)/(12 - 4) gradient_AB = -5/4

Find the midpoint, M, of AB: M = ((4 + 12)/2, (3 + (-7))/2) M = (8, -2)

Find the gradient of the perpendicular line, L: gradient_L = 4/5

Find the equation of line L through the midpoint M = (8, -2) with gradient 4/5: y - (-2) = 4/5(x - 8) or y + 2 = 4/5(x - 8)

The line parallel to $AB$ which passes through the point (5, 12) intersects $L$ at the point $C$. Find the coordinates of $C$.

The curves $y=x^{2}$ and $y^{2}=27x$ intersect at $O$(0, 0) and at the point $A$. Find the equation of the perpendicular bisector of the line $OA$.

The line $y=5x+6$ meets the curve $xy=8$ at the points $A$ and $B$.

Find the coordinates of $A$ and of $B$.

Extract:

The line y = 5x + 6 meets the curve xy = 8 at the points A and B.

Substituting the equation of the line into the equation of the curve and solving these equations simultaneously by substitution x(5x + 6) = 8

Expanding the brackets and setting the equation equal to 0 5x² + 6x - 8 = 0

Factorising to find the solutions (5x - 4)(x + 2) = 0 x = 4/5, x = -2

Substituting to find the y-coordinates y = 5(4/5) + 6 = 10 y = 5(-2) + 6 = -4

(4/5, 10) (-2, -4)

Find the coordinates of the point where the perpendicular bisector of the line $AB$ meets the line $y=x$.