, 2015

The $x$-intercept of a line occurs at the point whose $y$-coordinate is 0. Substituting $y=0$ into the equation of line 1, we get $0=2x+6$ or $2x=-6$ and so $x=-3$. Line 1 has $x$-intercept $-3$. (Point $P$ has coordinates $(-3,0)$.) Solution 1 The equation of line 2 with slope $-3$ and $y$-intercept $b$ is $y=-3x+b$. Line 2 passes through $Q(3,12)$ and so $x=3$ and $y=12$ satisfies the equation of line 2. Substituting, we get $12=-3(3)+b$ or $12=-9+b$ and so $b=21$. The equation of line 2 is $y=-3x+21$. Solution 2 A line with slope $m$ and passing through the point $(x_1,y_1)$ has equation $y-y_1=m(x-x_1)$. Since line 2 has slope $m=-3$ and passes through $Q(3,12)$, the slope-point equation of line 2 is $y-12=-3(x-3)$. We find the coordinates of point $R$ by substituting $y=0$ into the equation of line 2. This gives $0=-3x+21$ or $3x=21$ and so $x=7$. That is, line 2 has $x$-intercept $7$ (point $R$ has coordinates $(7,0)$). If we let the base of $\triangle PQR$ be side $PR$, then the height of the triangle is the vertical distance from $Q$ to $PR$. Since $Q$ has $y$-coordinate 12, then this height is 12. The $x$-coordinate of $P$ is $-3$ and the $x$-coordinate of $R$ is 7. (Both $y$-coordinates are 0.) Therefore, $PR$ has length $7-(-3)=10$. Finally, the area of $\triangle PQR$ is $\frac12\times10\times12=60$.