Straight Line flash cards

1. 1.Find the distance of a point $(x_1,y_1)$ from the line $ax+by+c=0$.

   $d=\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$

   Hint: Use the perpendicular distance formula.

2. 2.What is the angle $\theta$ between two lines with slopes $m_1$ and $m_2$?

   $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$; lines are parallel if $m_1=m_2$ and perpendicular if $m_1m_2=-1$.

   Hint: Think slope difference over $1+$ product.

3. 3.Write the equation of a line in intercept form and normal (perpendicular) form.

   Intercept form: $\dfrac{x}{a}+\dfrac{y}{b}=1$; Normal form: $x\cos\alpha+y\sin\alpha=p$, where $p$ is the perpendicular distance from the origin to the line.

   Hint: $a,b$ are the x- and y-intercepts; $\alpha$ is the angle the normal makes with the x-axis.

4. 4.Give the equation of a line through $(x_1,y_1)$ with slope $m$, and through two points $(x_1,y_1)$, $(x_2,y_2)$.

   Point-slope form: $y-y_1=m(x-x_1)$; Two-point form: $y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

   Hint: Slope of the line through the two points is $\dfrac{y_2-y_1}{x_2-x_1}$.

5. 5.How do you find the equations of the two lines that bisect the angles between the lines $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$?

   $\dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$ (the $+$ sign gives one bisector, the $-$ sign gives the other, and together they are always perpendicular to each other).

   Hint: Equate the perpendicular distances from a point on the bisector to each line, with both signs.

6. 6.If the family of lines $(2+\lambda)x+(1-3\lambda)y-5(1+\lambda)=0$ passes through a fixed point for every value of $\lambda$, find that point.

   Group terms in $\lambda$: $(2x+y-5)+\lambda(x-3y-5)=0$. This holds for all $\lambda$ only if both brackets vanish simultaneously: $2x+y=5$ and $x-3y=5$. Solving gives $y=-\dfrac{5}{7}$ and $x=\dfrac{20}{7}$. So every line of the family passes through $\left(\dfrac{20}{7},-\dfrac{5}{7}\right)$.

   Hint: Split the equation into two brackets, one free of $\lambda$.

7. 7.Find the value of $k$ for which the lines $x+2y-3=0$, $3x+ky-1=0$ and $2x-y+5=0$ are concurrent.

   First find the intersection of $x+2y-3=0$ and $2x-y+5=0$: solving gives $x=-\dfrac{7}{5}, y=\dfrac{11}{5}$. For concurrency this point must satisfy the third line: $3\left(-\dfrac{7}{5}\right)+k\left(\dfrac{11}{5}\right)-1=0 \Rightarrow -21+11k-5=0$, so $k=\dfrac{26}{11}$.

   Hint: Intersect the two lines without $k$ first, then substitute.

8. 8.Find the equation of the line passing through the intersection of $x+y-1=0$ and $x-y+3=0$, and also through the point $(2,3)$.

   The required line has the form $(x+y-1)+\lambda(x-y+3)=0$. Substituting $(2,3)$: $(2+3-1)+\lambda(2-3+3)=0 \Rightarrow 4+2\lambda=0 \Rightarrow \lambda=-2$. Substituting back, $(x+y-1)-2(x-y+3)=0$ simplifies to $x-3y+7=0$.

   Hint: Use the one-parameter family through the two lines' intersection.

9. 9.Prove that the lines $ax+by+c=0$, $bx+cy+a=0$, and $cx+ay+b=0$ (with $a,b,c$ not all equal) are concurrent whenever $a+b+c=0$.

   If $a+b+c=0$, test the point $(1,1)$ in each line: the first gives $a(1)+b(1)+c=a+b+c=0$, and the same sum, just re-ordered, vanishes for the other two equations as well. So $(1,1)$ satisfies all three lines simultaneously, proving they are concurrent at $(1,1)$ whenever $a+b+c=0$.

   Hint: Try the point $(1,1)$ directly in all three equations.

10. 10.The lines $3x-4y+5=0$ and $7x-8y+5=0$ intersect at a point $P$. Find $c$ so that the line $4x+5y+c=0$ also passes through $P$.

    Solving the first two equations (eliminate $y$ using twice the first minus the second) gives $x=5$, and back-substitution gives $y=5$, so $P=(5,5)$. For concurrency, $4(5)+5(5)+c=0 \Rightarrow 45+c=0$, so $c=-45$.

    Hint: Find the common point of the first two lines, then substitute.

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