CAIE IGCSE Mathematics Extended E3.7 Perpen…Lines

Covered syllabus points

E3.7.1 Find the gradient and equation of a straight line perpendicular to a

[Maximum number: 4]

16 A is the point (7,2) and B is the point (-5,8).

Find the equation of the line that is perpendicular to AB and that passes through the point (-1,3). Give your answer in the form y=m x+c.

y=2x+5 final answer

M1 for gradient of $AB=\frac{8-2}{-5-7}$ oe.
M1 for gradient of perpendicular, $p=-\frac{1}{\text{their gradient}}$ oe.
M1 dep for substituting (-1,3) into $y=\text{their }px+c$ , where their $p\ne0$ .

[Maximum number: 7]

9 A is the point (0,2), B is the point (3,3) and C is the point (4,0).

(a)

Find the equation of the perpendicular bisector of AC.
Give your answer in the form y=mx+c.
y=

[ 4 ]

y=2x-3
M1 for $\frac{-1}{\text{their grad }(b)(i)}$
B1 for (2,1)
M1 for substituting their (2,1) into y= their px+d oe

(b)

ABCD is a kite.

The point D has coordinates (w, 4 w+1).
Find the coordinates of D.

[ 3 ]

(-2,-7)
B2 for w=-2
or M1 for 4w+1=2w-3 FT their (b)(ii) or for $2=\frac{4w+1-3}{w-3}$

[Maximum number: 6]

Figure for Question 3 — CAIE IGCSE Mathematics Extended

O is the origin (0,0), A is the point (8,1) and B is the point (2,5).

(a)

Find the equation of the perpendicular bisector of AB. Give your answer in the form y=m x+c.

y=

[ 4 ]

$y=\frac32x-\frac92$ oe
B1 for (5,3) oe
M1 for gradient $=-\frac1{\text{their gradient of }AB}$
M1 substituting their midpoint into y= their mx+c

(b)

The line AB meets the y-axis at P.

The perpendicular bisector of AB meets the y-axis at Q.
Find the length of PQ.

[ 2 ]

$\frac{65}{6}$ oe
M1 for their $\frac{19}{3}$ - their $-\frac92$ oe

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CAIE IGCSE Mathematics Extended E3.7 Perpendicular Lines Topic Practice

Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)