"Coordinate Geometry" — Practice

Question 1

Find the distance between the parallel lines $3x-4y+7=0$ and $3x-4y-13=0$.

Accepted Answer

Both lines have the same $a$ and $b$, so the gap is $\dfrac{\lvert c_1-c_2\rvert}{\sqrt{a^{2}+b^{2}}}$.

Here that is $\dfrac{\lvert 7-(-13)\rvert}{\sqrt{9+16}}=\dfrac{20}{5}$.

The distance is **4**.

Question 2

Do the points $(2,3)$ and $(-1,1)$ lie on the same side of the line $x+2y-7=0$?

Accepted Answer

Evaluate $x+2y-7$ at each point, without taking the modulus.

At $(2,3)$: $2+6-7=+1$.

At $(-1,1)$: $-1+2-7=-6$.

The signs differ, so the points are on **opposite sides**. This sign test is faster than any sketch, and it is how you check whether a point lies inside a triangle — test it against all three sides.

Question 3

Find the line through the intersection of $2x-3y+4=0$ and $3x+4y-5=0$ that is parallel to $6x-7y+8=0$.

Accepted Answer

Use the family $(2x-3y+4)+\lambda(3x+4y-5)=0$, which is $(2+3\lambda)x+(4\lambda-3)y+(4-5\lambda)=0$.

Parallel to $6x-7y+8=0$ means the coefficients are proportional: $\dfrac{2+3\lambda}{4\lambda-3}=\dfrac{6}{-7}$.

Cross-multiplying, $-7(2+3\lambda)=6(4\lambda-3)$, so $-14-21\lambda=24\lambda-18$ and $45\lambda=4$, giving $\lambda=\dfrac{4}{45}$.

Substituting back and clearing the denominator by multiplying through by $45$:

**$102x-119y+160=0$**

Check the slope: $\dfrac{102}{119}=\dfrac{6}{7}$, which matches. Solving for the intersection first would have meant carrying the fractions $\left(-	frac{1}{17},	frac{22}{17}ight)$ through the whole question.

Question 4

Find the orthocentre of the triangle with vertices $A(1,2)$, $B(5,2)$ and $C(3,6)$.

Accepted Answer

$AB$ lies along $y=2$, so the altitude from $C$ is the vertical line $x=3$.

For a second altitude, $BC$ has slope $\dfrac{6-2}{3-5}=-2$, so the altitude from $A$ has slope $	frac12$: $y-2=	frac12(x-1)$.

At $x=3$ this gives $y-2=1$, so the orthocentre is **$(3,3)$**.

Confirm with the third altitude: $AC$ has slope $2$, so the altitude from $B(5,2)$ has slope $-	frac12$, giving $y-2=-	frac12(x-5)$, which at $x=3$ is again $y=3$.

Question 5

The line $3x+4y=12$ meets the axes at $A$ and $B$. Find the incentre of triangle $OAB$.

Accepted Answer

The intercepts give $A(4,0)$ and $B(0,3)$, with $O$ the origin. The sides are $OA=4$, $OB=3$ and $AB=5$.

The incentre is $\dfrac{a\vec{A}+b\vec{B}+c\vec{C}}{a+b+c}$, where each length is the side opposite that vertex. Opposite $A$ is $OB=3$, opposite $B$ is $OA=4$, opposite $O$ is $AB=5$.

$x=\dfrac{3(4)+4(0)+5(0)}{12}=1$ and $y=\dfrac{3(0)+4(3)+5(0)}{12}=1$.

The incentre is **$(1,1)$**, and since it sits one unit from both axes the inradius is $1$ — worth noting as a check, because the incentre of a right triangle is always at distance $r$ from both legs.

Question 6

A variable line through $(2,3)$ meets the coordinate axes at $P$ and $Q$. Find the locus of the midpoint of $PQ$.

Accepted Answer

Write the line as $\dfrac{x}{p}+\dfrac{y}{q}=1$, so $P(p,0)$ and $Q(0,q)$. Passing through $(2,3)$ forces $\dfrac{2}{p}+\dfrac{3}{q}=1$.

The midpoint is $(h,k)=\left(\dfrac{p}{2},\dfrac{q}{2}\right)$, so $p=2h$ and $q=2k$.

Substituting: $\dfrac{2}{2h}+\dfrac{3}{2k}=1$. Multiplying by $2hk$ gives $2k+3h=2hk$.

The locus is **$3x+2y=2xy$**.

The pattern is the standard one: write the moving point in terms of parameters, use the given condition to relate them, then eliminate.

Question 7

Find the centre and radius of $x^{2}+y^{2}-4x+6y-12=0$.

Accepted Answer

Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$ gives $2g=-4$ and $2f=6$, so $g=-2$ and $f=3$, and $c=-12$.

The centre is $(-g,-f)=$ **$(2,-3)$**.

The radius is $\sqrt{g^{2}+f^{2}-c}=\sqrt{4+9+12}=$ **$5$**.

Note both sign flips: the equation shows $-4x$ and the centre's $x$-coordinate is $+2$.

Question 8

Find the length of the tangent from $(5,4)$ to the circle $x^{2}+y^{2}-2x-4y-4=0$.

Accepted Answer

The tangent length is $\sqrt{S_1}$, where $S_1$ is the left-hand side evaluated at the point.

$S_1=25+16-10-16-4=11$.

The tangent length is **$\sqrt{11}$**.

That $S_1$ came out positive also confirms the point is outside the circle, which it had to be for a tangent to exist at all.

Question 9

Find the length of the chord cut off on $x^{2}+y^{2}=9$ by the line $3x+4y=5$.

Accepted Answer

The perpendicular from the centre bisects the chord, so half-chord, $d$ and $r$ are the sides of a right triangle and the chord is $2\sqrt{r^{2}-d^{2}}$.

Here $d=\dfrac{\lvert -5\rvert}{\sqrt{9+16}}=1$ and $r=3$.

Chord $=2\sqrt{9-1}=2\cdot 2\sqrt{2}=$ **$4\sqrt{2}$**.

Since $d=1<3=r$ the line genuinely cuts the circle, which the same single calculation already told you.

Question 10

Find the common chord of $x^{2}+y^{2}=25$ and $x^{2}+y^{2}-8x-6y+21=0$, and its length.

Accepted Answer

Subtracting, the $x^{2}$ and $y^{2}$ terms cancel: $-25-(-8x-6y+21)=0$, giving $8x+6y-46=0$, that is

**$4x+3y-23=0$** — the radical axis, and therefore the common chord.

For its length, use the first circle: $d=\dfrac{23}{5}$ and $r=5$, so the chord is $2\sqrt{25-\dfrac{529}{25}}=2\cdot\dfrac{\sqrt{96}}{5}=$ **$\dfrac{8\sqrt{6}}{5}$**.

Check it against the second circle, which has centre $(4,3)$ and radius $2$: its distance to the line is $\dfrac{2}{5}$, and $2\sqrt{4-\dfrac{4}{25}}=\dfrac{8\sqrt{6}}{5}$ again. Two circles, one chord, same answer.

Question 11

Find the circle whose centre lies on $2x+y=3$ and which passes through $(1,1)$ and $(2,-1)$.

Accepted Answer

Put the centre at $(h,\,3-2h)$, which satisfies the line automatically and leaves one unknown.

Equal distances to the two points: $(h-1)^{2}+(2-2h)^{2}=(h-2)^{2}+(4-2h)^{2}$.

Expanding, $5h^{2}-10h+5=5h^{2}-20h+20$, so $10h=15$ and $h=	frac32$. The centre is $\left(	frac32,0ight)$.

Then $r^{2}=\left(	frac32-1ight)^{2}+1=	frac54$, and the circle is

**$x^{2}+y^{2}-3x+1=0$**

Verify both points: $1+1-3+1=0$ and $4+1-6+1=0$.

Question 12

Find the focus, directrix and latus rectum of $y^{2}=-12x$.

Accepted Answer

$4a=-12$, so $a=-3$. The negative sign simply means the parabola opens to the left.

Focus $(a,0)=$ **$(-3,0)$**; directrix $x=-a$, that is **$x=3$**; latus rectum $\lvert 4a\rvert=$ **$12$**.

Sanity check the geometry: the focus is inside the curve and the directrix on the other side of the vertex, which is what the two signs are telling you.

Question 13

Find the eccentricity and the foci of $9x^{2}+4y^{2}=36$.

Accepted Answer

Dividing by $36$ gives $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}=1$.

The larger denominator is under $y$, so the major axis is **vertical** — this is the step most answers get wrong. So the semi-major axis squared is $9$ and the semi-minor squared is $4$.

$e^{2}=1-\dfrac{4}{9}=\dfrac{5}{9}$, so $e=$ **$\dfrac{\sqrt5}{3}$**.

The foci lie on the major axis, at $(0,\pm 3e)=$ **$(0,\pm\sqrt5)$**.

Question 14

For what values of $k$ does the line $y=3x+k$ touch the ellipse $\dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1$?

Accepted Answer

The tangency condition for an ellipse is $c^{2}=a^{2}m^{2}+b^{2}$, which is the unified $c^{2}=Am^{2}+B$ with $A=a^{2}$ and $B=b^{2}$.

Here $m=3$, $a^{2}=16$, $b^{2}=9$, so $k^{2}=16(9)+9=153$.

**$k=\pm 3\sqrt{17}$**

Two values, because a line of given slope touches an ellipse on each side of it.

Question 15

Two tangents are drawn from $(4,3)$ to $\dfrac{x^{2}}{9}-\dfrac{y^{2}}{4}=1$. Find the product of their slopes.

Accepted Answer

A tangent is $y=mx+c$ with $c^{2}=a^{2}m^{2}-b^{2}=9m^{2}-4$.

Passing through $(4,3)$ forces $c=3-4m$, so $(3-4m)^{2}=9m^{2}-4$.

Expanding, $9-24m+16m^{2}=9m^{2}-4$, giving $7m^{2}-24m+13=0$.

The two roots are the two tangent slopes, so their product is **$\dfrac{13}{7}$** (and their sum is $\dfrac{24}{7}$).

You never had to find either slope. Setting this product to $-1$ instead is exactly how the director circle is derived.

Question 16

Perpendicular tangents are drawn to $\dfrac{x^{2}}{25}+\dfrac{y^{2}}{16}=1$. Show that their point of intersection lies on a circle, and find its radius.

Accepted Answer

From $(h,k)$, a tangent has $c=k-mh$, and the condition $c^{2}=Am^{2}+B$ gives $(k-mh)^{2}=Am^{2}+B$.

Collecting in $m$: $m^{2}(h^{2}-A)-2hk\,m+(k^{2}-B)=0$.

The product of the roots is $\dfrac{k^{2}-B}{h^{2}-A}$, and perpendicularity sets it to $-1$, so $k^{2}-B=A-h^{2}$, that is $h^{2}+k^{2}=A+B$.

With $A=25$ and $B=16$ the locus is the director circle **$x^{2}+y^{2}=41$**, of radius **$\sqrt{41}$**.

The same derivation gives $2a^{2}$ for a circle and $a^{2}-b^{2}$ for a hyperbola — one calculation, three results.

Practice — "Coordinate Geometry"

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