Straight Lines Test 38

Given,

Abscissa = Ordinate

Let the coordinate of the variable point P is (x, y)

Now, the abscissa of this point = xand its ordinate = y

According to the question,

x = y

⇒ x – y = 0

So, the locus of the point is x – y = 0

The equation of the line passing through \(\left(x_{1}, y_{1}\right)\) with the slope \(m\) is:

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

The slope of line passing through the points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is:

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

The slope of line passing through the points \((2,5)(-3,6)\) is \(=\frac{6-5}{-3-2}\)

\(\Rightarrow m_{1}=\frac{1}{-5}=\frac{-1}{5}\)

The two non-vertical lines are perpendicular to each other only if and only if their slopes are negative reciprocals of each other.

The slope of perpendicular to the line through the points \((2,5)(-3,6)\) is \(\mathrm{m}_{2}=\frac{-1}{\mathrm{~m}_{1}}\)

\(\Rightarrow \mathrm{m}_{2}=5\)

The equation of the line passing through \((x_{1}, y_{1})=(-3,5)\) and perpendicular to the line through the points \((2,5)(-3,6)\) is:

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

\(\Rightarrow(y-5)=5(x+3)\)

\(\Rightarrow 5 x-y+20=0\)

Given,

Point \((1,2)\) divides the line passing through \((h, 0)\) and \((0, k)\) in the ratio \(2: 3\)

Here, \(m: n=2: 3\)

\((x, y)=(1,2),\left(x_{1}, y_{1}\right)=(0, k),\left(x_{2}, y_{2}\right)=(h, 0)\)

Using section formula,

\(x=\frac{\operatorname{mx}_{x}+\mathrm{nx}_{1}}{\mathrm{~m}+\mathrm{n}}\)

\(\Rightarrow 1=\frac{2(\mathrm{~h})+3(0)}{2+3}\)

\(\Rightarrow 1=\frac{2 \mathrm{~h}}{5}\)

\(\Rightarrow 2 \mathrm{~h}=5\)

\(\Rightarrow \mathrm{h}=\frac{5}{2}\) 

\(\mathrm{y}=\frac{\mathrm{m} y_{2}+\mathrm{n} y_{1}}{\mathrm{~m}+\mathrm{n}}\)

\(\Rightarrow 2=\frac{2(0)+3(\mathrm{k})}{2+3}\)

\(\Rightarrow 3 \mathrm{k}=2(5)=10\)

\(\Rightarrow \mathrm{k}=\frac{10}{3}\)

Points are \(\left(\frac{5}{2}, 0\right)\) and \(\left(0, \frac{10}{3}\right)\)

Now, equation of line passing through \(\left(\frac{5}{2}, 0\right)\) and \(\left(0, \frac{10}{3}\right)\)

\(\frac{x}{\frac{5}{2}}+\frac{y}{\frac{10}{3}}=1\)

\(\Rightarrow \frac{2 x}{5}+\frac{3 y}{10}=1\)

\(\Rightarrow 4 x+3 y=10\)    (because multiply whole equation by 10)

We know, distance between two points \(\left(x_1, y_1\right)\) and \(\left(x_2, y_2\right)\) is \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\).

So, distance between \((5,12)\) from origin \((0,0)\) is \(\sqrt{(5-0)^2+(12-0)^2}=\sqrt{(5)^2+(12)^2}=13\) units.

Given,

The length of the perpendicular from the point \((4,1)\) on the line \(3 x-4 y+k=0\) is 2 units.

The perpendicular distance \(d\) from \(P\left(x_{1}, y_{1}\right)\) to the line \(a x+b y+c=0\) is given by

\(d=\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|\).........(i)

Let \(P=(4,1)\)

Here \(x_{1}=4, y_{1}=1, a=3, b=-4\) and \(d=2\)

Now substitute \(x_{1}=4\) and \(y_{1}=1\) in the equation (i) we get,

\(\Rightarrow d=\left|\frac{3 × x_{1}-4 × y_{1}+k}{\sqrt{3^{2}+(-4)^{2}}}\right|=2\)

\(\Rightarrow d=\left|\frac{3 × 4-4 × 2+k}{\sqrt{25}}\right|=2\)

\(\Rightarrow|\frac{8+\mathrm{k}}{5}|=2\)

\(\Rightarrow|8+\mathrm{k}|\) = \(10\)

\(\Rightarrow \mathrm{k}=2 \text { or }-18\)

Given,

\(x+2 y=5\)...(i)

\(3 x+7 y=17\)...(ii)

Solving equation (i) and (ii), we get

\(x=1 \text { and } y=2\)

Point of intersection: \((x, y)=\left(x_{1}, y_{1}\right)=(1,2)\)

Let slope of the straight line \(3 x+4 y=10\) is \(m_{1}\)

Slope \(\left(m_{1}\right)=\frac{-3}{4}\)

We know that when two lines are perpendicular, the product of their slope is \(-1 .\)

Slope of perpendicular line \(m=\frac{-1}{m_{1}}=\frac{4}{3}\)

Equation of line: \(\left(y-y_{1}\right)=m\left(x-x_{1}\right)\)

\(\Rightarrow y-2=\frac{4}{3}(x-1)\)

\(\Rightarrow 3 y-6=4 x-4 \)

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

Given,

Line \(x-3 y+5=0\)

The general equation of a line is \(y=m x+c\)

Where \(m\) is the slope and \(c\) is any constant

• The slope of parallel lines is equal.
• The slope of the perpendicular line have their product \(=-1\)

\(\Rightarrow y=\frac{1}{3} x+\frac{5}{3}\)

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

Now for the slope of the perpendicular line \(m_2\)

\(m_{1} \times m_{2}=-1\)

\(\Rightarrow \frac{1}{3} \times m_{2}=-1\)

\(\Rightarrow m_{2}=-3\)

Perpendicular line has the slope \(-3\) and passes through \((2,-4)\)

\(\therefore\) Equation of the perpendicular line is

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

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

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

Let \(L_{3}\) is the line passing through the intersection of lines \(L_{1}\) and \(L_{2}\) Then equation of line \(L_{3}\) is given by \(L_{1}+(\lambda) L_{2}=0\)

Equation of line passing through the intersection of lines, \(2 x-3 y+7=0\) and \(7 x+4 y+2=0\) is given by,

\(2 x-3 y+7+\lambda(7 x+4 y+2)=0\)

Now, this line is also passing through the point \((2,3)\),

So, it will satisfy the equation of line \(2 x-3 y+7+\lambda(7 x+4 y+2)=0\)

\(\therefore 2(2)-3(3)+7+\lambda(7(2)+4(3)+2)=0\)

\(\Rightarrow 4-9+7=-\lambda(14+12+2) \)

\(\Rightarrow-\lambda=\frac{2}{28} \)

\(\Rightarrow \lambda=-\frac{1}{14}\)

Now, putting \(\lambda=-\frac{1}{14}\) in \((1)\), we get equation of line,

\(2 x-3 y+7+\left(-\frac{1}{14}\right)(7 x+4 y+2)=0\)

Multiply by 14 ,

\(\Rightarrow 28 x-42 y+98-(7 x+4 y+2)=0\)

\(\Rightarrow 28 x-7 x-42 y-4 y+98-2=0 \)

\(\Rightarrow 21 x-46 y+96=0\)

Given,

The distance of the point \((k+2,2 k+3)\) from the line \(x+3 y=7\) is \(\frac{4}{\sqrt{10}}\).

Let \(P=(k+2,2 k+3)\)

\(\Rightarrow x_{1}=k+2\) and \(y_{1}=2 k+3\)

Here, \(a=1\) and \(b=3\)

Now substitute \(x_{1}=k+2\) and \(y_{1}=2 k+3\) in the equation \(x+3 y-7=0\) we get,

\(\Rightarrow\left|x_{1}+3 \cdot y_{1}-7\right|=|7 k+4|\)

\(\Rightarrow \sqrt{a^{2}+b^{2}}\)

\(=\sqrt{1^{2}+3^{2}}\)

\(=\sqrt{10}\)

As we know that, the perpendicular distance \(d\) from \(P\left(x_{1}, y_{1}\right)\) to the line \(a x+b y+c\) \(=0\) is given by:

\(d=\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|\)

\( d=\left|\frac{7 k+4}{\sqrt{1^{2}+3^{2}}}\right|\)

\(=\frac{4}{\sqrt{10}} \)

\(\Rightarrow|7 k+4|\)

\(=4\)

\(\Rightarrow k=0 \text { or }\frac{-8}{ 7}\)

Given,

Angle between the lines \(x+y-3=0\) and \(x-y+3=0\) is \(\alpha\).

Angle between the lines \(x -\sqrt{3} y+2 \sqrt{3}=0\) and \(\sqrt{3} x-y+1=0\) is \(\beta\).

The angle \(\theta\) between the lines having slope \(m_{1}\) and \(m_{2}\) is given by \(\tan \theta=\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|\)

Let's find the slope,

Slope of line \(x+y-3=0\) is \(m_{1}\) and slope of line \(x-y+3=0\) is \(m_{2}\).

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

Therefore,

tan \(\mathrm{a}=\left|\frac{1-(-1)}{1+(-1) \times 1}\right|=\infty\)

\(\alpha=90^{\circ}\)

Slope of line \(x-\sqrt{3} y+2 \sqrt{3}=0\) is \(\mathrm{m}_{1}\) and \(\sqrt{3} x-y+1=0\) is \(m_{2}\)

\( \mathrm{m}_{1}=(\frac{1 }{\mathrm{N}})\) and \(\mathrm{m}_{2}=\sqrt{3}\)

\(\tan \beta=\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\left(\sqrt{3} \times \frac{1}{\sqrt{3}}\right)}\right|=\frac{1}{\sqrt{3}}\)

\(\beta=30^{\circ}\)

\(\alpha>\beta\)