Answer:
Step-by-step explanation:
[tex]\text{9. Solution:}\\\text{Let the L2 denote the line of whose equation is need to be found.}\\\text{We have,}\\y=-4x+5..........(1)\\\text{Comparing equation(1) with }y=m_1x+b,\\\text{Slope of line(1), }m_1=-4\\\text{Since the line L2 is parallel to the line(1), }\\\text{Slope of line L2 = Slope of line 1 }\\\text{or, }m_2=-4[/tex]
[tex]\text{Now, L2 passes through (-3,-1). Therefore, it's equation will be:}\\y-(-1)=-4\{x-(-3)\}\\\text{or, }y+1=-4(x+3)\\\text{or, }y+1=-4x-12\\\text{or, }y=-4x-13\text{ is the required equation.}[/tex]
[tex]\text{10. Solution:}\\\text{Let }m_2\text{ be the slope of the line whose equation is need to be found.}\\\text{We have another line,}\\y=-3x+0.75\\\text{Comparing with }y=mx+b,\\\text{Slope }(m_1)=-3\\\text{Using the condition of perpendicular lines,}\\m_1.m_2=-1\\\text{or, }-3.m_2=-1\\\text{or, }m_2=1/3\\[/tex]
[tex]\text{Now, the equation of the line having slope }m_2\text{ and that passes through the }\\\text{point }(x_1,y_1) \text{ is:}\\y-y_1=m_2(x-x_1)\\\text{In this case, }m_2=\dfrac{1}{3}\text{ and }(x_1,y_1)=(1,-5).\\\therefore\ y+5=\dfrac{1}{3}(x-1)\text{ is the required equation.}[/tex]
[tex]\text{11. Solution,}\\\text{The equation of the line passing through the points }(x_1,y_1)\text{ and }(x_2,y_2)\text{ is}\\\text{given by:}\\y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\\text{According to the question, }\\(x_1,y_1)=(60,20)\\(x_2,y_2)=(220,120)\\\text{So the equation of AB is:}\\y-20=\dfrac{120-20}{220-60}(x-60)\\\\\text{or, }y-20=\dfrac{100}{160}(x-60)\\\\\text{or, }y-20=\dfrac{5}{8}(x-60)\\\\\text{or, }8y-160=5(x-60)\\\text{or, }8y-160=5x-300\\\text{or, }5x-8y=140[/tex]
