CAIE IGCSE Additional Math 7 Straight Line Graphs

Covered syllabus points

7.1. Use straight-line equations 7.2. Use parallel and perpendicular gradients 7.3. Solve midpoint, length and bisector problems 7.4. Transform relationships to straight-line form

[Maximum number: 2]

point A has coordinates ( -2,4 ).
The point B has coordinates (6,10).
The point C has coordinates (12,2).

Find the gradients of the lines A B, A C and B C.

Grad $A B=\frac{6}{8}$ oe

Grad $A C=-\frac{2}{14}$ oe

Grad $B C=-\frac{8}{6}$ oe B1 for 2 correct unsimplified

[Maximum number: 10]

The coordinates of points A and B are (-5,6) and (4,-6) respectively. The point C lies on the line AB, between A and B, such that $\frac{AC}{CB}=\frac12$ .

(a)

The line CD is perpendicular to AB. Find the equation of CD in the form y=mx+c.

[ 4 ]

$m_{AB}=\frac{-6-6}{4-(-5)}=-\frac43$ $m_{CD}=\frac34$ $y-2=\frac34(x+2)$ $y=\frac34x+\frac72$

(b)

The length of BD is $\sqrt{125}$ . Find the coordinates of the two possible positions of point D.

[ 6 ]

$(x-4)^2+(y+6)^2=125$
Use $y=\frac34x+\frac72$ to eliminate one unknown
$(x-4)^2+\left(\frac34x+\frac72+6\right)^2=125$ $25x^2+100x-300=0$
Factorises or solves the quadratic
$(2,5)\quad\text{and}\quad(-6,-1)$

[Maximum number: 7]

4 The coordinates of points A, B, C and D are as follows.

A(-4,3)\quad B(6,-9)\quad C(15,10)\quad D(14,-1)

The line L has equation y=11 x-75 .

The perpendicular bisector of the line A B meets L at the point E .

Find the area of triangle C D E .

$m_{A B}=\frac{-9-3}{6--4}$ oe isw or $-\frac{6}{5}$ $\left[m_{\perp}=\right] \frac{5}{6}$ or $\frac{-1}{\text { their } m_{A B}}$
FT their $m_{A B}$ providing that
$\frac{\text { difference in } y}{\text { difference in } x}$ attempted
[Midpoint] $\left(\frac{-4+6}{2}, \frac{3-9}{2}\right)$ oe isw or (1,-3)
Finds equation of perpendicular bisector:
$y \text {-their }(-3)=-\frac{1}{\text { their }\left(-\frac{6}{5}\right)}(x-\text { their } 1) \mathrm{oe}$
FT their perpendicular gradient providing gradient is $\frac{-1}{\text { their } m_{A B}}$ and using their midpoint

Their midpoint must not be any of A, B, C or D $11 x-75=\frac{5}{6} x-\frac{23}{6}$
Eliminates one variable
dep on previous $\mathbf{M}$ mark
11 x-75= their perp bisector equation
E(7,2)
[Area C D E= ] 40
Dependent on all previous marks

[Maximum number: 4]

The coordinates of points A and B are (-5,6) and (4,-6) respectively. The point C lies on the line AB, between A and B, such that $\frac{AC}{CB}=\frac12$ .

The line CD is perpendicular to AB. Find the equation of CD in the form y=mx+c.

$m_{AB}=\frac{-6-6}{4-(-5)}=-\frac43$ $m_{CD}=\frac34$ $y-2=\frac34(x+2)$ $y=\frac34x+\frac72$

[Maximum number: 4]

Variables x and y are such that when $\ln y$ is plotted against x, a straight-line graph is obtained.
The line passes through the points $(1,\ln 15)$ and $(2,\ln 75)$ .
Show that $y=Ab^x$ , where A and b are integers to be found.

$\ln y=mx+c$ soi
Correct method to find m, e.g. $m=\frac{\ln 75-\ln 15}{2-1}$ oe, or $-m=\ln 15-\ln 75$ oe, or $m=\ln 5$ oe.
M1 FT their c if necessary.
Correct method to find c, e.g. $\ln 15=(\text{their }\ln 5)(1)+c$ oe, $\ln 75=(\text{their }\ln 5)(2)+c$ oe, or $c=\ln 3$ oe, soi.
M1 FT their m if necessary.
$y=3\times 5^x$ nfww, mark final answer

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CAIE IGCSE Additional Math 7 Straight Line Graphs Topic Practice

Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question 4

Question 5