# Test: Coordinate Geometry II - Ambitious

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Question 1:   Rewrite the equation of a straight line in the slope-intercept form: $3x-5y+2=0$
$y=\frac{3}{5}x+\frac{2}{5}$
$y=\frac{5}{3}x+\frac{5}{2}$
$y=\frac{5}{3}x-\frac{5}{2}$
$y=\frac{3}{5}x+2$
Question 2:   Rewrite the equation of a straight line in the standard form: $y=-3\frac{1}{2}x+2$
$x+7y+4=0$
$7x+2y-4=0$
$1x+y-4=0$
$3x-y+2=0$
Question 3:   Find the equation of a straight line going through the points $A\left(12,2\right)$ and $B\left(1,-1\right)$
$y=\frac{3}{11}x-1\frac{3}{11}$
$y=\frac{2}{7}x-\frac{3}{5}$
$y=\frac{5}{8}x-1\frac{2}{5}$
$y=\frac{7}{9}x-1\frac{3}{7}$
Question 4:   Find the equation of a straight line going through the points $A\left(-3,7\right)$ and $B\left(7,2\right)$
$y=-\frac{3}{4}x+6\frac{1}{3}$
$y=-\frac{3}{7}x+4\frac{2}{3}$
$y=-\frac{2}{3}x+3\frac{1}{5}$
$y=-\frac{1}{2}x+5\frac{1}{2}$
Question 5:   Find the equation of a straight line knowing that it crosses $x$ - axis at $-3.5$ and $y$ -axis at $7$
$y=x+7$
$y=2x+7$
$y=x+3.5$
$y=7x+3.5$
Question 6:   Find where the line $y=-\frac{1}{4}x+10$ crosses the $x$ -axis
$\left(20,0\right)$
$\left(40,0\right)$
$\left(4,0\right)$
$\left(10,0\right)$
Question 7:   Find where the two lines intersect: Line one: $y=-\frac{1}{2}x-5\frac{1}{2}$ Line two: $y=-x+6$
$\left(21,-16\right)$
$\left(18,-13\right)$
$\left(23,-17\right)$
$\left(11\frac{1}{2},-16\right)$
Question 8:   Find where the two lines intersect: Line one: $y=-3x+4$ Line two: $y=2x-5$
$\left(2,-1\frac{1}{3}\right)$
$\left(2\frac{1}{5},-\frac{4}{5}\right)$
$\left(1\frac{3}{4},-1\frac{1}{5}\right)$
$\left(1\frac{4}{5},-1\frac{2}{5}\right)$
Question 9:   Which of the below is the equation of a straight line parallel to: $y=1\frac{1}{4}x-4$
$4x-5y+5=0$
$5x-4y+9=0$
$5x+4y+12=0$
$4x+5y+10=0$
Question 10:   Which of the below is the equation of a straight line perpendicular to: $y=\frac{3}{4}x+1\frac{1}{4}$
$3x-4y-4=0$
$4x-3y-2=0$
$4x+3y-6=0$
$3x+4y-8=0$