Linear Graphs y = mx + c: past paper questions & worked solutions

Use change in y divided by the corresponding change in x.

The line \[ y=2-5x \] has gradient \(-5\).

A parallel line has the same gradient, so line L has form \[ y=-5x+c \]

Line L passes through \((0,6)\), so \[ 6=c \]

So the gradient of line P is \(-\frac25\).

A line perpendicular to P has gradient \[ \frac52 \]

The line passes through \((0,-3)\) and \((2,0)\).

The y-intercept is \(-3\), so \[ y=\frac32x-3 \]

The midpoint is \((a,19)\).

For the x-coordinate: \[ a=\frac{3+47}{2}=25 \]

For the y-coordinate: \[ 19=\frac{11+b}{2} \]

Rearrange into the form y=mx+c.

The gradient is \(\frac52\).

The coefficient of x is the gradient.

Every point on the y-axis has x-coordinate zero.

The line passes through \((0,1)\) and \((2,5)\).

Line \(L\) has equation

So its gradient is

Line \(M\) is perpendicular to \(L\), so its gradient is

Line \(M\) passes through \((8,1)\).

From the graph, the line crosses the y-axis at \(1\), so \[ c=1 \]

It also crosses the x-axis at \(2\), so using \((0,1)\) and \((2,0)\), \[ m=\frac{0-1}{2-0}=-\frac12 \]

Therefore \[ y=-\frac12x+1 \]

Using point \((-5,12)\): \[ y-12=-\frac52(x+5) \]

Multiply by 2: \[ 2y-24=-5x-25 \]

From the graph, the line crosses the y-axis at \(-1\), so \[ c=-1 \]

Using two points on the line, for example \((0,-1)\) and \((2,2)\), the gradient is \[ \frac{2-(-1)}{2-0}=\frac32 \]

So the equation is \[ y=\frac32x-1 \]

So the gradient is \[ -\frac72 \]

For the x-axis, \(y=0\): \[ 2(0)+7x=10 \]

(a) The gradient of \(AB\), where \(A(5,8)\) and \(B(9,-4)\), is \[ \frac{-4-8}{9-5}=\frac{-12}{4}=-3 \]

(b) A line perpendicular to a line with gradient \(-3\) has gradient \[ \frac13 \]

Useful points are: \[ (-2,15),\quad (0,7),\quad (1,3),\quad (2,-1),\quad (3,-5) \]

Plot these points and join them with a straight line.

\(y=7+4x\) has gradient \(4\), so \(\mathbf L\) has gradient \(-\frac14\).

Its gradient is \(-2\), so a perpendicular line has gradient \(\frac12\).

Line \(L\) passes through \((8,11)\): \[ y-11=\frac12(x-8) \]

The gradient of \(AB\) is

Since \(ABCD\) is a square, \(BC\) is perpendicular to \(AB\).

So the gradient of \(BC\) is

The line through \(B=(3,5)\) is

Multiply by 3:

The distance between the centres is \[ \sqrt{(7-(-1))^2+(18-3)^2} \]

The radii are \(3\) cm and \(15\) cm.

Since the distance between the centres is \(17\) cm, which is less than the sum of the radii, the circles intersect.

The centre of the circle is

The point \(A\) is

Since \(AOP\) is a diameter, \(O\) is the midpoint of \(AP\).

Let \(P=(x,y)\).

The gradient of \(OP\) is

The tangent is perpendicular to the radius, so its gradient is

Using point \(P=(2,3)\),

Line \(L\) passes through \((4,-1)\) and \((6,4)\).

Line \(M\) is perpendicular to \(L\), so its gradient is \[ -\frac25 \]

Line \(M\) intersects the y-axis at \((0,8)\), so \[ y=-\frac25x+8 \]

At the x-axis, \(y=0\): \[ 0=-\frac25x+8 \]

(a) The midpoint of \(A(5,-4)\) and \(B(13,1)\) is \[ \left(\frac{5+13}{2},\frac{-4+1}{2}\right) \]

The gradient is \(-3\).

(c) For \(x=100\), \[ y=2-3(100)=-298 \]

The point has \(y=-302\), not \(-298\), so it does not lie on the line.

In a rhombus, the diagonals are perpendicular and bisect each other.

The diagonal \(AC\) lies on

So the gradient of \(AC\) is

Therefore the gradient of \(BD\) is

The line \(BD\) passes through

So

To find where this line intersects the \(y\)-axis, put \(x=0\):

Diagonals of a rhombus bisect each other at right angles.

Midpoint of \(AC\):

Gradient of \(AC\):

So gradient of \(BD\) is \(-\frac13\).

The gradient of diagonal \(PR\) is

The other diagonal \(QS\) is perpendicular to \(PR\), so its gradient is

The diagonals of a square bisect each other.

The midpoint of \(PR\) is

So \(QS\) passes through \((6,1)\) with gradient \(\frac13\).

Multiply by 3:

The tangent point lies on the tangent line, so

The tangent gradient is \(-\frac54\), so the radius gradient is \(\frac45\).

Using centre \((3q,4q)\) and tangent point \((1,4)\),

So the centre is \((6,8)\). Therefore

The diagonals of a square bisect each other and are perpendicular.

Midpoint of \(P(4,7)\) and \(R(8,-5)\):

Gradient of \(PR\):

So the perpendicular gradient is \(\frac13\).

The line through \(Q\) and \(S\) passes through \((6,1)\), so

So

Substitute into the curve:

Divide by 5:

Factorise:

So

Using \(y=5-x\):

So

Substitute into

Factorise:

So

Using \(x=5-2y\):

So

Substitute into

Factorise:

So

Using \(y=2x+1\):

So \[ x=2y-1 \]

So \[ y=14 \quad \text{or} \quad y=-1 \]

If \(y=14\), then \(x=27\).

If \(y=-1\), then \(x=-3\).

The gradient of \(AC\) is \(-\frac67\).

Since \(\angle ABC=90^\circ\), lines \(AB\) and \(BC\) are perpendicular.

The product of perpendicular gradients is \(-1\):

Now solve

Double the second equation:

Subtract:

Then

Since \(AB=AC\), point \(A\) lies on the perpendicular bisector of \(BC\).

The midpoint of \(BC\) is

The gradient of \(BC\) is

So the gradient of the perpendicular bisector is

Use the line through \(M\):

Multiply by 10:

Divide by 2:

A parallel line has the same gradient, \(-4\). One possible answer is \[ y=-4x \]

(b) The gradient of line \(L\) through \((-3,1)\) and \((2,-2)\) is \[ \frac{-2-1}{2-(-3)}=-\frac35 \]

A perpendicular line has gradient \(\frac53\).

Through \((-6,4)\): \[ y-4=\frac53(x+6) \]

Multiply by 3: \[ 3y-12=5x+30 \]

Substitute this into the circle equation:

So

Using \(y=x+2\):

For \(L_1\),

So the gradient of \(L_1\) is \(-\frac12\).

Since \(L_2\) is perpendicular to \(L_1\), the gradient of \(L_2\) is \(2\).

Using the point \((8,6)\):

At \(A\), the line crosses the \(x\)-axis, so \(y=0\):

So \(A=(5,0)\).

At \(B\), \(x=-3\):

So \(B=(-3,-16)\).

Triangle \(AOB\) has base \(OA=5\) and perpendicular height \(16\).

Point \(A=(0.2,2.6)\) lies on the line.

So

Use point \(A=(0.2,2.6)\):

Now substitute \(y=3x+2\) into the curve:

One root is \(0.2=\frac15\).

The product of the roots is

So the other root is

Then

The length of \(PQ\) is \(4\sqrt{10}\), so

Square both sides:

Since \(a>0\),

So

The midpoint of \(PQ\) is

The gradient of \(PQ\) is

So the gradient of the perpendicular bisector is

Using point \((1,4)\):

The centre of the circle is \((0,0)\), and

The gradient of \(OP\) is

So the gradient of the tangent is

The tangent through \(P\) is

At the \(x\)-axis, \(y=0\):

So

At the \(y\)-axis, \(x=0\):

Therefore

In the kite, \(AB=AD\) and \(CB=CD\), so \(AC\) is the line of symmetry.

Therefore \(D\) is the reflection of \(B\) in the line \(AC\).

A direction vector for \(AC\) is

Let \(H\) be the foot of the perpendicular from \(B\) to \(AC\).

The fraction along \(AC\) to \(H\) is

So

Since \(H\) is the midpoint of \(BD\),

so

The gradient of the perpendicular bisector is \(2\), so the gradient of \(AB\) is

Since \(A=(j,7)\) and \(B=(6,k)\),

The midpoint of \(AB\) is

Substitute \(j=2k-8\):

The midpoint lies on \(y=2x+7\), so

Then

Therefore the midpoint is

Line \(L\) is \[ x-y=3 \]

So \[ y=x-3 \]

Substitute \(y=x-3\): \[ 3x-(x-3)+(x-3)^2=9 \]

So \(x=1\) or \(x=3\).

If \(x=1\), then \(y=-2\). If \(x=3\), then \(y=0\).

The midpoint of \((1,-2)\) and \((3,0)\) is \[ \left(\frac{1+3}{2},\frac{-2+0}{2}\right)=(2,-1) \]

Since \(M\) is the midpoint of \(JK\),

The gradient of \(HM\) is 2, and \(H=(-4,1)\).

Also,

So

Substitute \(k=2j+15\):

Since \(j<0\),

Then

The gradient of \(BD\) is

So \(B=(-2,1)\) and \(D=(8,7)\).

In this kite, \(AC\) is the perpendicular bisector of \(BD\).

The gradient of \(BD\) is \(\frac35\), so the gradient of \(AC\) is

Using point \((3,4)\):

The midpoint of \(BD\) is \(M=(0,2)\).

Since \(A=(-3,4)\), the gradient of \(AM\) is

For the kite, diagonal \(AC\) is perpendicular to \(BD\), so the gradient of \(BD\) is

Use direction vector \((2,3)\) along \(BD\).

and

Since \(AB=6.5\),

The \(x\)-coordinate of \(B\) is positive, so use \(t=1.5\).

The midpoint of \(AB\) is

The gradient of \(AB\) is

So the perpendicular gradient is

Line \(M\) passes through \(\left(-\frac12,\frac52\right)\), so

At point \(C\), \(x=-1\):

So

Now use vectors from \(A\):

Area of triangle \(ABC\) is

The vector \(AB\) is \[ (1-(-4),10-(-2))=(5,12) \]

The vector \(BC\) is \[ (19-1,a-10)=(18,a-10) \]

Adjacent sides of a rectangle are perpendicular, so their dot product is zero: \[ 5(18)+12(a-10)=0 \]

Now \[ BC=(18,-7.5) \]

So \[ D=A+BC=(-4,-2)+(18,-7.5)=(14,-9.5) \]

Perimeter: \[ 2(13+19.5)=65 \]

The line passes through \((0,4)\) and \((2,0)\).

The \(y\)-intercept is \(4\), so

For \(y=3x-5\), put \(x=0\):

So the crossing point is \((0,-5)\).