1. Find the equation of the straight line passing through points (1, 4) and (3, 5). The slope $$m= \frac{5-4}{3-1}=\frac{1}{2}$$. Using point-slope form with point (1,4): $$y-4=\frac{1}{2}(x-1) \Rightarrow 2y-8=x-1 \Rightarrow x-2y+7=0$$. Answer is (c). 2. Find the x-intercept of the line $$-e + f x - g y = h$$. Set $$y=0$$: $$-e + f x = h \Rightarrow f x = e + h \Rightarrow x=\frac{e+h}{f}$$. So x-intercept is $$\left(\frac{e+h}{f},0\right)$$. Answer is (b). 3. Find the slope of the line passing through points $$(-2a,4b)$$ and $$(4b,-2a)$$. Slope $$m=\frac{-2a - 4b}{4b - (-2a)}=\frac{-2a-4b}{4b+2a} = -1$$ (since numerator and denominator are negatives of each other). Answer is (c). 4. Find equation of line with x-intercept $-4$ and y-intercept $5$. Equation: $$\frac{x}{-4}+ \frac{y}{5}=1 \Rightarrow 5x -4y = 20 \Rightarrow 5x - 4y -20 =0$$. Answer is (c). 5. Find equation of line through point $$(-1,3)$$ and intersection of $$\ell_1:5x -3y -17=0$$ and $$\ell_2:-2x + 5y + 22=0$$. Solve for intersection: Multiply $$\ell_1$$ by 5 and $$\ell_2$$ by 3: $$25x -15y -85=0$$ $$-6x +15y +66=0$$ Add: $$19x -19=0 \Rightarrow x=1$$. Substitute in $$\ell_1$$: $$5(1) -3y -17=0 \Rightarrow -3y = 12 \Rightarrow y=-4$$. Line through points $$(-1,3)$$ and $$(1,-4)$$ slope $$m= \frac{-4-3}{1+1} = \frac{-7}{2}$$. Using point-slope form: $$y-3 = -\frac{7}{2}(x+1) \Rightarrow 2(y-3) = -7(x+1) \Rightarrow 2y-6 = -7x -7 \Rightarrow 7x + 2y + 1=0$$. Answer is (d). 6. Find equation of line through $$(-1,1)$$ parallel to $$3x + 4y - 7=0$$. Slope of given line $$m = -\frac{3}{4}$$. Using point-slope form: $$y-1 = -\frac{3}{4}(x+1) \Rightarrow 4(y-1) = -3(x+1) \Rightarrow 4y -4 = -3x -3 \Rightarrow 3x + 4y -1=0$$. Answer is (c). 7. Find equation of line through $$(5,-2)$$ perpendicular to $$x - 2y - 4=0$$. Slope of given line $$m= \frac{1}{2}$$. Perpendicular slope $$m_p = -2$$. Using point-slope form: $$y + 2 = -2(x -5) \Rightarrow y = -2x + 10 - 2 = -2x + 8$$. Answer is (b). 8. Find value of $$A$$ s.t. line through $$(A, 4)$$ and $$(10, A)$$ is parallel to $$6x + 5y = 35$$. Slope of given line $$m = -\frac{6}{5}$$. Slope of line through points: $$m = \frac{A - 4}{10 - A}$$. Set equal: $$\frac{A - 4}{10 - A} = -\frac{6}{5} \Rightarrow 5(A-4) = -6(10-A) \Rightarrow 5A -20 = -60 +6A \Rightarrow 5A - 6A = -60 + 20 \Rightarrow -A = -40 \Rightarrow A = 40$$. Answer is (a). 9. Find $$A$$ given point $$(2,5)$$ lies on $$-2x + A y = 15$$. Substitute: $$-2(2) + A(5) = 15 \Rightarrow -4 + 5A = 15 \Rightarrow 5A = 19 \Rightarrow A= \frac{19}{5}$$. Answer is (a). 10. For what $$k$$ does the system $$\begin{cases} 2x -3y = -27 \\ 5x - ky = 4 \end{cases}$$ have no solution? For no solution, lines must be parallel but distinct: Check ratio: $$\frac{2}{5} = \frac{-3}{-k} \Rightarrow \frac{2}{5} = \frac{3}{k} \Rightarrow 2k = 15 \Rightarrow k = \frac{15}{2}$$. Answer is (d).