Exercises: Intersection of two lines

Line equation of $f$

In order to determine the line equation of $f$, first read off two points from the diagram that lie on the straight line $f$. In the above example, you may for instance get $A(0|3)$ and $B(4|2)$. Use these two points to determine the slope of $f$ via the difference quotient

$${\displaystyle m_f=\frac{y_B-y_A}{x_B-x_A}}$$

Plug in the values

$${\displaystyle m_f=\frac{2-3}{4-0}=-\frac{1}{4}}$$

Now determine the $y$-axis intercept $b_f$ by plugging in any point on $f$ into the general line equation. Alternatively, you may directly read off at which value $f$ intersects the $y$-axis (if possible).

$f(x):y=m_{f}x+b_f$

For instance, plug in $A$.

$${\displaystyle 3=-\frac{1}{4}\cdot 0+b_f}$$

Simplify.

$3=b_f\Rightarrow b_f=3$

So the line equation is $${\displaystyle f(x)=-\frac{1}{4}\cdot x+3}$$.

Line equation of $g$

In order to determine the line equation of $g$, first read off two points from the diagram that lie on the straight line $g$. In the above example, you may for instance get $C(-4|0)$ and $D(0|1)$. Use these two points to determine the slope of $g$ via the difference quotient

$${\displaystyle m_g=\frac{y_D-y_C}{x_D-x_C}}$$

Plug in the values

$${\displaystyle m_g=\frac{1-0}{0-(-4)}=\frac{1}{4}}$$

Now determine the $y$-axis intercept $b_g$ by plugging in any point on $g$ into the general line equation. Alternatively, you may directly read off at which value $g$ intersects the $y$-axis (if possible).

$g(x):y=m_{g}x+b_g$

For instance, plug in $D$.

$${\displaystyle 1=\frac{1}{4}\cdot 0+b_g}$$

Simplify.

$${\displaystyle 1=b_g\Rightarrow b_g=1}$$

So the line equation is $${\displaystyle g(x)=\frac{1}{4}\cdot x+1}$$.

Line equation of $h$

In order to determine the line equation of $h$, first read off two points from the diagram that lie on the straight line $h$. In the above example, you may for instance get $E(-1|0)$ and $A(0|3)$. Use these two points to determine the slope of $h$ via the difference quotient

$${\displaystyle m_h=\frac{y_A-y_E}{x_A-x_E}}$$

Plug in the values

$${\displaystyle m_h=\frac{3-0}{0-(-1)}=3}$$

Now determine the $y$-axis intercept $b_h$ by plugging in any point on $h$ into the general line equation. Alternatively, you may directly read off at which value $h$ intersects the $y$-axis (if possible).

$h(x):y=m_{h}x+b_h$

For instance, plug in $A$.

$3=3\cdot 0+b_h$

Simplify.

$3=b_h\Rightarrow b_h=3$

So the line equation is $${\displaystyle h(x)=3\cdot x+3}$$.

Line equation of $i$

In order to determine the line equation of $g$, first read off two points from the diagram that lie on the straight line $g$. In the above example, you may for instance get $F(0|-3)$ and $S(6|0)$. Use these two points to determine the slope of $g$ via the difference quotient

$${\displaystyle m_i=\frac{y_S-y_F}{x_S-x_F}}$$

Plug in the values

$${\displaystyle m_i=\frac{0-(-3)}{6-0}=\frac{1}{2}}$$

Now determine the $y$-axis intercept $b_i$ by plugging in any point on $i$ into the general line equation. Alternatively, you may directly read off at which value $i$ intersects the $y$-axis (if possible).

$i(x):y=m_{i}x+b_i$

For instance, plug in $F$.

$${\displaystyle -3=\frac{1}{2}\cdot 0+b_i}$$

Simplify.

$-3=b_i\Rightarrow b_i=-3$

So the line equation is $${\displaystyle i(x)=\frac{1}{2}\cdot x-3}$$.

Intersection point $P(x_p|y_p)$ of $g$ and $h$

To determine the point of intersection of two functions, you set them equal and solve for $x$. The function equations (determined in the previous problem) are $$g(x):y={\displaystyle \frac{1}{4}x+1}$$ and $h(x):y=3x+3$.

Now substitute $-\frac{8}{11}$ into the line equation of $g$ or $h$ to determine $y_P$.

$h(x_P):y_P=3\cdot x_P+3$

Plug in $x_P$ .

$${\displaystyle y_P=3\cdot (-\frac{8}{11})+3=\frac{9}{11}}$$

The lines $g$ and $h$ therefore intersect at $${\displaystyle P(-\frac{8}{11}|\frac{9}{11})}$$.

You may obtain the zero $x_{N_f}$ of $f$ by setting the function equation $${\displaystyle f(x):y=-\frac{1}{4}x+3}$$ equal to 0 and solving for $x$.

So $f$ has the zero 12.

The intersection of $h$ and $i$ and the intersection of $g$ and $i$ lie outside the figure above.

Intersection point $T(x_T|y_T)$ of $h$ and $i$

To determine the point of intersection of two functions, you set them equal and solve for $x$. The function equations (determined in the previous problem) are $h(x):y=3x+3$ and $${\displaystyle i(x):y=\frac{1}{2}x-3}$$.

Now substitute $-\frac{12}{5}$ into the line equation of $h$ or $i$ to determine $y_T$.

$h(x_T):y_T=3\cdot x_T+3$

Plug in $x_T$.

$${\displaystyle y_T=3\cdot (-\frac{12}{5})+3=-\frac{21}{5}}$$

The lines $h$ and $i$ therefore intersect at $${\displaystyle T(-\frac{5}{2}|-\frac{21}{5})}$$.

Intersection point $Q(x_Q|y_Q)$ of $g$ and $i$

To determine the point of intersection of two functions, you set them equal and solve for $x$. The function equations (determined in the previous problem) are $${\displaystyle g(x):y=\frac{1}{4}x+1}$$ and $${\displaystyle i(x):y=\frac{1}{2}x-3}$$.

Now substitute $16$ into the line equation of $g$ or $i$ to determine $y_Q$.

$${\displaystyle g(x_Q):y_Q=\frac{1}{4}\cdot x_Q+1}$$

Plug in $x_Q$.

$${\displaystyle y_Q=\frac{1}{4}\cdot 16+1=5}$$

The lines $g$ and $i$ therefore intersect at $${\displaystyle Q(16|5)}$$.

If no pair of lines is parallel, then there are a total of 6 intersections, namely the following:

• $f$ and $g$

• $f$ and $h$

• $f$ and $i$

• $g$ and $h$

• $g$ and $i$

• $h$ and $i$