Straight Lines Test 36

Given,

The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150 degrees with the positive direction of the \(x\)-axis.

We know that, If \(p\) is the length of the normal from origin to a line and \(\alpha\) is the angle made by the normal with the positive direction of \(x\)-axis then the equation of the line is given by:

\(x \cos \alpha+y \sin \alpha=p\)

Now, the equation of a line is

\(x \times \cos 150^{\circ} +y \times \sin 150^{\circ} =7\)

\(\Rightarrow x \times \cos (180^{\circ}-30^{\circ})+y \times \sin (180^{\circ}-30^{\circ}) =7\)\(\quad\quad(\because cos (180^{\circ}-\theta) = cos \theta,sin (180^{\circ}-\theta) = sin \theta)\)

\(\Rightarrow x \times \cos 30^{\circ}+y \times \sin 30^{\circ}=7\)

\(\Rightarrow \frac{\sqrt{3} x}{2}+\frac{y}{2}=7\) \(\quad(\because \cos 30^{\circ}=\frac{\sqrt{3}}{2} \text { and } \sin 30^{\circ}\)=\(\frac{1}{2})\)

\(\Rightarrow \sqrt{3} x+y=7 \times 2\)

\(\Rightarrow \sqrt{3} x+y=14\)

Given,

\(3 x+y=7\)

\(\Rightarrow y=-3 x+7\)

Then the slope of a line perpendicular

The slope of line \(=m=-3\)

Then the slope of a line perpendicular to it is \(\frac{-1}{m}=\frac{1}{3}\)

The equation of line passing through \((1,1)\) with slope \(\frac{1}{3}\) is:

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

Then the slope of a line perpendicular

\(\Rightarrow 3 y-3=x-1\)

Then the slope of a line perpendicular

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

Then the slope of a line perpendicular

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

Then the slope of a line perpendicular

Then the slope of a line perpendicular

For \(x\)-intercept, \(y=0\)

\(x=-2\)

Then the slope of a line perpendicular

\(\therefore\)The \(x\)-intercept of line is \(-2\)

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 \((1,3)\) and \((5,6)\) is \(\sqrt{(1-5)^2+(3-6)^2}=\sqrt{(4)^2+(3)^2}=5\) units.

Given,

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

\(\sqrt3 y-x+6=0\)

\(y-\sqrt{3} x-5=0\)

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

So, slope of line, \(m_{1}=\sqrt{3}\)

\(\sqrt{3} y-x+6=0\)

\(\Rightarrow y=\frac{x}{\sqrt{3}}-\frac{6}{\sqrt{3}}\)

So, slope of the line, \(m_{2}=\frac{1}{\sqrt{3}}\)

Let \(\theta\) be the acute angle between the lines.

The angle \(\theta\) between the lines having slope \(m_{1}\) and \(m_{2}\) is given by:

\(\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|\)

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

\(\Rightarrow \tan \theta=\left|\frac{\frac{2}{\sqrt{3}}}{2}\right| \)

\(\Rightarrow \tan \theta=\frac{1}{\sqrt{3}} \)

\(\Rightarrow \theta=30^{\circ}\)

Given,

The distances of \(\mathrm{P}(x, y)\) from \(\mathrm{A}(4,1)\) and \(\mathrm{B}(-1,4)\) are equal.

The distance ' \(d\) ' between two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is obtained by using the Pythagoras' Theorem: \(d^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}\).

Using the formula for the distance between two points:

\(\mathrm{AP}^{2}=(x-4)^{2}+(y-1)^{2}\)

\(\mathrm{BP}^{2}=(x+1)^{2}+(y-4)^{2}\)

Since the distances are equal, we have:

\(\mathrm{AP}^{2}=\mathrm{BP}^{2} \)

\(\Rightarrow(x-4)^{2}+(y-1)^{2}=(x+1)^{2}+(y-4)^{2}\)

\(\Rightarrow x^{2}-8 x+16+y^{2}-2 y+1=x^{2}+2 x+1+y^{2}-8 y+16 \)

\(\Rightarrow 10 x=6 y\)

\(\Rightarrow 5 x=3 y\)

Let \(P(h, k)\) be any point on the locus which is equidistant from the point A(1, 3) and B(-2, 1).

According to question,

\(P A=P B\)

\(\sqrt{(h-1)^{2}+(k-3)^{2}}=\sqrt{(h+2)^{2}+(k-1)^{2}}\)..(i)

Squaring both side in equation (i) we get,

\(\Rightarrow(h-1)^{2}+(k-3)^{2}=(h+2)^{2}+(k-1)^{2} \)\(\quad\quad(\because (a\pm b)^{2} = a^{2}+b^{2} \pm 2ab)\)

\(\Rightarrow h^{2}-2 h+1+k^{2}-6 k+9=h^{2}+4 h+4+k^{2}-2 k+1\)

\(\Rightarrow-2 h-6 k+10=4 h-2 k+5 \)

\(\Rightarrow 6 h+4 k=5\)

\(\therefore\)The locus of \((h,k)\) is \(6 x+4 y=5\).

Given,

The area of a triangle is 5 square units.

Equation of line is \(5 x-y=0\)

The equation of a line perpendicular to a given line is

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

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

\(\Rightarrow \frac{x}{\lambda}+\frac{y}{\left(\frac{\lambda}{5}\right)}=1\)

Area of triangle \(=5\) square units

Area of triangle = \(\frac{1}{2}×b×h\)

\(\frac{1}{2} \times \lambda \times \frac{\lambda}{5}=5\)

\(\Rightarrow \lambda^{2}=50 \)

\(\Rightarrow\lambda=\pm 5 \sqrt{2}\)

Put the value of \(\lambda\) in equation (i) we get,

 \(x+5 y=\pm 5 \sqrt{2}\)

\(\Rightarrow x+5 y \pm 5 \sqrt{2}=0\)

Given,

The slope is zero.

Let θ be the angle of inclination of the given line with the positive direction of x-axis in the anticlockwise sense.

Then its slope is given by:

m = tan θ

tan θ = 0

⇒ θ = 0°

⇒ Either the line is x-axis or it is parallel to the x-axis.

Given,

Two parallel line\(p(x+y)+q=0\) and \(p(x+y)-r=0\)

The distance between the parallel lines \(ax+by+c_{1}=0\) and \(a x+b y+c_{2}=0\) is given by:

\(d=\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|\)

Here, we have to find the distance between the parallel lines \(p(x+y)+q=0\) and \(p(x\) \(+y)-r=0\)

The given equations of line can be re-written as: \(p x+p y+q=0\) and \(p x+p y-r=0\)

By comparing the equations of the given line with \(ax+by+\mathrm{c}_{1}=0\) and \(ax+by+\) \(c_{2}=0\) we get,

\(a=p, b=p, c_{1}=q\) and \(c_{2}=-r\)

As we know that, the distance between the parallel lines is given by:

\(d=\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|\)

\(d=\left|\frac{q+r}{\sqrt{p^{2}+p^{2}}}\right|\)

\(=\frac{|q+r|}{\sqrt{2} p}\)

Distance between the parallel lines \(p(x+y)+q=0\) and \(p(x+y)-r=0\) is\(\frac{|q+r|}{\sqrt{2} p}\).

Let \(P(h, k)\) be any position of the moving point and let \(A(a, 0)\) and \(B(-a, 0)\) be the given points. Then

\( P A^2+P B^2=2 c^2\)

\( \Rightarrow(h-a)^2+(k-0)^2+(h+a)^2+(k-0)^2=2 c^2\)

\( \Rightarrow h^2-2 a h+a^2+k^2+h^2+2 a h+a^2+k^2=2 c^2 \)

\( \Rightarrow 2 h^2+2 k^2+2 a^2=2 c^2\)

\( \Rightarrow h^2+k^2+a^2=c^2\)

\( \Rightarrow h^2+k^2=c^2-a^2\)

Hence, the locus of \((h, k)\) is \(x^2+y^2=c^2-a^2\)