Straight Lines Test 37

As we know that, the equation of a line whose slope is \(m\) and which makes an intercept \(c\) on the \(y\) axis is given by:

\(y=m x+c\)

Let the slope of the required line be \(m\).

Here, we have \(c=-2\)

So, the equation the required line is \(y=mx-2\).....(i)

The required line passes through the point \((-1,5)\)

Therefore,

Substitute \(x=-1\) and \(y=5\) will satisfy the equation (i).

\( 5=-m-2\)

\(\Rightarrow m=-7\)

\(\therefore\)The equation of the required line is \(7 x+y+2=0\).

Given, the point \((2,3)\) and slope of the line is \(2\)

By, slope-intercept formula,

\(y-3=2(x-2)\)

\(\Rightarrow y-3=2 x-4\)

\(\Rightarrow 2 x-4-y+3=0\)

\(\Rightarrow 2 x-y-1=0\)

The coordinates of a point dividing the line segment joining \(\left(x_1, y_1\right)\) and \(\left( x _2, y _2\right)\) externally in the ratio \(m : n\) is \(\left(\frac{m x_2-n x_1}{m-n}, \frac{m y_2-n y_1}{m-n}\right)\).

So, the coordinates of a point dividing the line segment joining \((1,2)\) and \((4,5)\) externally in the ratio \(2: 1\) is \(\left(\frac{2 \times 4-1 \times 1}{2-1}, \frac{2 \times 5-1 \times 2}{2-1}\right)=(7,8)\).

\(\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1\)

Or, \(b x\cos \theta+a y \sin \theta-a b=0\).......(1)

Length of the perpendicular from point \(\left(\sqrt{a^{2}-b^{2}} ; 0\right)\) to line (1) is

\(P_{1}=\frac{|b \cos \theta\left(\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b \mid}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{\mid b \cos \theta \sqrt{a^{2}-b^{2}}-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)....(2)

Length of the perpendicular from point \(\left(-\sqrt{a^{2}-b^{2}}, 0\right)\) to line \((2)\) is

\(P_{2}=\frac{|b \cos \theta\left(-\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|b \cos \theta (-\sqrt{a^{2}-b^{2}})-a b \mid}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)

\(P_{2}=\frac{|b \cos \theta\left(-\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|b \cos \theta \sqrt{a^{2}-b^{2}}+a b \mid}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)....(3)

On multiplying equations (2) and (3) we get,

\(P_{1} P_{2}=\frac{|(b \cos \theta \sqrt{a^{2}-b^{2}}-a b)(b \cos \theta \sqrt{a^{2}-b^{2}}+a b) \mid}{\left(\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\right)^{2}}\)

\(=\frac{\left|\left(b \cos \theta \sqrt{a^{2}-b^{2}}-a b\right)\left(b \cos \theta \sqrt{a^{2}-b^{2}}+a b\right)\right|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}\)

\(=\frac{\left|\left(b \cos \theta \sqrt{a^{2}-b^{2}}\right)^{2}-(a b)^{2}\right|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}\)\(\quad(\because (a+b)(a-b) = a^{2} -b^{2})\)

\(=\frac{\left|b^{2} \cos ^{2} \theta\left(a^{2}-b^{2}\right)-a^{2} b^{2}\right|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}\)

\(=\frac{\left|a^{2} b^{2} \cos ^{2} \theta-b^{4} \cos ^{2} \theta-a^{2} b^{2}\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\)

\(=\frac{b^{2}\left|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2}\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\)

\(=\frac{b^{2}\left|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2} \sin ^{2} \theta-a^{2} \cos ^{2} \theta\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\) \(\quad\quad\because[\sin ^{2} \theta+\cos ^{2} \theta=1]\)

\(=\frac{b^{2}\left|-\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\)

\(=\frac{b^{2}\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}\)

\(=b^{2}\)

Given,

The inclination of the line joining the points \((x,-3)\) and \((2,5)\) is \(135^{\circ}\)

Here, we have to find the value of \(x\).

As we know that, the slope of a line is given by tan \(\alpha\) and it is denoted by \(m\) where \(\alpha\) \(\neq \frac{\pi}{2}\) and it represents the angle which a given line makes with respect to the \(X-\) axis in the positive direction.

\(\mathrm{m}=\tan \left(135^{\circ}\right)=\tan \left(180^{\circ}-135^{\circ}\right)\)

\(=-\tan 45^{\circ}\)

\(=-1\)

As we know that, the slope of the line joining the points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is:

\(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Here, \(x_{1}=x, y_{1}=-3, x_{2}=2, y_{2}=5\) and \(m=-1\).

\(-1=\frac{5-(-3)}{2-x}\)

\(\Rightarrow-2+x=8\)

\(\Rightarrow x=10\)

Given,

The line passing through the points \((3,2)\) and \((1,-1)\)

Equation of a line passing through \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is:

\(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Therefore,

\(\frac{y-2}{x-3}=\frac{-1-2}{1-3} \)\(\quad\quad(\because \frac{y-2}{x-3}=\frac{-1-2}{1-3} )\)

\(3 x-2 y-5=0\)

\(y=\frac{3}{2} x-\frac{5}{2} \)

\(\Rightarrow \text { Slope }\left(m_{1}\right)=\frac{3}{2} \text { and } c_{1}=\frac{-5}{2}\)

Now for the slope of the perpendicular line \(\left(\mathrm{m}_{2}\right)\)

\(\mathrm{m}_{1} \times \mathrm{m}_{2}=-1 \)

\(\Rightarrow \frac{3}{2} \times \mathrm{m}_{2}=-1 \)

\(\Rightarrow \mathrm{m}_{2}=\frac{-2}{3}\)

Given,

Lines \(x-y+4=0\)...(i)

\(y-2 x-5=0\)...(ii)

The equation of the line with points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\)

\(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Adding the 2 equation (i) and (ii)

\(-x-1=0\)

\(x=-1\)

Putting it in equation (i) we get,

\(-1-y+4=0 \)

\(y=3\)

Intersection of the lines \((-1,3)\)

Now the equation of the line to be find out is

\(\frac{y-3}{x-(-1)}=\frac{2-3}{3-(-1)}\)

\(4(y-3)=-1(x+1)\)

\(x+4 y-11=0\)

Given,

Points are \(\mathrm{A}(-1,0)\) and \(\mathrm{B}(0,5)\).

The distance between tow points \(\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) and \(\mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)\) is given by:

\(\mathrm{AB}=\sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}}\)

Let the point which is equidistant from the points \(\mathrm{A}\) and \(\mathrm{B}\) be \(\mathrm{C}(\mathrm{x},\mathrm{x})\).

As the point is equidistant we have \(\mathrm{A}\mathrm{C}=\mathrm{B}\mathrm{C}\).

Therefore, equating the right hand and left hand side using distance formula.

\(\mathrm{AC} =\mathrm{BC} \)

\(\sqrt{(-1-\mathrm{x})^{2}+(0-\mathrm{x})^{2}}=\sqrt{(0-\mathrm{x})^{2}+(5-\mathrm{x})^{2}}\)

\(\Rightarrow (-1-\mathrm{x})^{2}+(0-\mathrm{x})^{2} =(0-\mathrm{x})^{2}+(5-\mathrm{x})^{2}\)

\(\Rightarrow1+2 \mathrm{x}+\mathrm{x}^{2}+\mathrm{x}^{2} =\mathrm{x}^{2}+25-10 \mathrm{x}+\mathrm{x}^{2} \)

\(\Rightarrow1+2 \mathrm{x} =25-10 \mathrm{x}\)

\(\Rightarrow12 \mathrm{x} =24 \)

\(\Rightarrow\mathrm{x} =2\)

Thus, the point which is equidistant from both points is \((2,2)\).

Given,

Straight-line passing through \((5,-2)\) and \((-4,7)\)

The slope of a line passing through the distinct points \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) is given by,

\(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Where, \(x_{1}\) = 5, \(y_{1}\) = -2, \(x_{2}\) = -4, and \(y_{2}\) = 7

Slope of a line \(=m=\frac{7-(-2)}{-4-5}\)

\(=\frac{9}{-9}\)

\(=-1\)

Now, equation of line is:

\((y-y_{1})=m(x-x_{1})\)

\(y-(-2)=-1 \times(x-5)\)

\(\Rightarrow y+2=-x+5\)

\(\Rightarrow x+y=3\)

Given,

Slope of line = -2

Lines are:

\(2 x-y=1 \)....(i)

\(x+2 y=3\)...(ii)

On multiplying equation (i) by 2 we get,

\(4 x-2y=2 \)....(iii)

Adding equation (ii) and (iii) we get,

\(5 x=5\)

\(\Rightarrow x=1\)

On putting value of \(x\) in equation (i),

\(2(1)-y=1 \)

\(\Rightarrow y=1\)

Therefore,

The intersection point is \((1,1)\)

So line has the slope m = \(-2\) and passes through \((1,1)\)

Therefore,

Equation of the perpendicular line is:

\(\left(y-y_{1}\right)=m\left(x-x_{1}\right)\)

\(\Rightarrow y-1=-2(x-1) \)

\(\Rightarrow y+2 x-3=0\)