Straight Lines Test -9

Since the lines cut off intercepts – 3 on y-axis then the line is passing through the point (0, - 3).

Given that: \(\tan \theta = \frac{3}{5}\)

⇒ Slope of the line m = \(\frac{3}{5}\)

So, the equation of the line is

\(y – y_1 = m(x – x_1)\)

⇒ \(y+3 = \frac{3}{5}(x - 0)\)

⇒ 5y + 15 = 3x

⇒ 3x – 5y – 15 = 0

⇒ 5y – 3x + 15 = 0

Intercept form of a line is

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

⇒ \(\frac{x}{a} + \frac{y}{a} = 1 (\because a = b)\)

⇒ x + y = a

⇒ y = - x + a

\(\therefore \) Slope is – 1

Equation of any line perpendicular to the given

line x + y + 1 = 0 is x – y + k = 0 …(i)

If eq. (i) passes through the point (1, 2) then

1 – 2 + k = 0 ⇒ k = 1

Putting the value of k in eq. (i) we have

x – y + 1 = 0 ⇒ y – x – 1 = 0

First equation of line having intercepts on the axes

a, - b is \(\frac{x}{a} - \frac{y}{b} = 1\)

⇒ bx – ay = ab …(i)

Second equation of line having intercepts on the axes

b, - a is \(\frac{x}{b} - \frac{y}{a} = 1\)

⇒ ax – by = ab …(ii)

Slope of eq. (i) \(m_1 = \frac{b}{a}\)

Slope of eq. (ii) \(m_2 = \frac{a}{b}\)

\(\therefore tan \theta = |\frac{m_1 - m_2}{1 + m_1m_2}|\)

\(= \frac{\frac{b}{a} - \frac{a}{b}}{1 + \frac{a}{b}\frac{b}{a}} = \frac{b^2 - a^2}{2ab}\)

Equation of line passing through the points (2, - 3) and (4, - 5) is

\(y + 3 = \frac{-5 + 3}{4 - 2}(x-2)\)

⇒ y + 3 = \(\frac{-2}{2}(x - 2)\)

⇒ y + 3 = - (x – 2)

⇒ y + 3 = - x + 2

⇒ x + y = - 1

⇒ \(\frac{x}{-1} + \frac{y}{-1} = 1\) (intercept form)

\(\therefore\) a = - 1, b = - 1

Given equation are:

2x – 3y + 5 = 0 …(i)

3x + 4y = 0 …(ii)

From eq. (ii) we get,

4y = - 3x ⇒ y = \(\frac{-3}{4}x \dots(iii)\)

Putting the value of y in eq. (i) we have

\(2x - 3(\frac{-3}{4}x) + 5 = 0\)

⇒ 8x + 9x + 20 = 0

⇒ 17x + 20 = 0

⇒ \(x = \frac{-20}{17}\)

Putting the value of x in eq. (iii) we get

\(y = \frac{-3}{4}(\frac{-20}{17})\)

⇒ \(y = \frac{15}{17}\)

\(\therefore\) Point of intersection is \((-\frac{20}{17}, \frac{15}{17})\)

Now perpendicular distance from the point \((-\frac{20}{17}, \frac{15}{17})\) to the given line 5x – 2y = 0 is

\(|\frac{5(-\frac{20}{17}) - 2(\frac{15}{17})}{\sqrt{25 + 4}}| \\ = |\frac{\frac{-100}{7} - \frac{30}{17}}{\sqrt {29}}|\)

\(= \frac{130}{17\sqrt {29}}\)

Equation of line is given by

\(\sqrt3 x + y + 1 = 0\)

⇒ y = -\(\sqrt 3x - 1\)

\(\therefore \) Slope of this line, \(m_1 = -\sqrt 3\)

Let \(m_2\) be the slope of the required line

\(\therefore tan \theta = |\frac{m_1 - m_2}{1 + m_1m_2}| \\ \implies tan 60^o = |\frac{-\sqrt3 - m_2}{1 + (-\sqrt3 m_2)}|\)

\(\implies \sqrt3 = \pm(\frac{-\sqrt3 - m_2}{1 - \sqrt3 m_2}) \)

⇒ \(\sqrt 3 = (\frac{-\sqrt3 - m_2}{1 - \sqrt3 m_2})\) [taking (+) sign]

⇒ \(\sqrt 3 - 3m_2 = -\sqrt3 - m_2\)

⇒ \(2 m_2 = 2\sqrt 3 \implies m_2 = \sqrt3\)

and \(\sqrt 3 = \frac{\sqrt3 + m_2}{1 - \sqrt 3 m_2}\) [taking (-) sign]

⇒ \(\sqrt3 - 3m_2 = \sqrt3 + m_2\)

⇒ \( 4m_2 = 0 \implies m_2 = 0\)

\(\therefore\) Equation of line passing through (3, - 2) with slope \(\sqrt3\) is

\(y + 2 = \sqrt3(x-3)\)

⇒ \(y+2 = \sqrt3x - 3\sqrt3\)

⇒ \(\sqrt3x - y - 2 - 3\sqrt3 = 0\)

And the equation of line passing through (3, - 2) with slope 0 is

y + 2 = 0(x – 3) ⇒ y + 2 = 0

Equation of any line passing through (1, 0) is

y – 0 = m(x – 1) ⇒ mx – y – m = 0

Distance of the line from origin is \(\frac{\sqrt3}{2}\)

\(\therefore \frac{\sqrt 3}{2} = |\frac{m \times 0 - 0 - m}{\sqrt{1 + m^2}}|\)

⇒ \(\frac{\sqrt3}{2} = |\frac{-m}{\sqrt{1 + m^2}}|\)

Squaring both sides, we get

\(\frac{3}{4} = \frac{m^2}{1 + m^2}\)

\(\implies 4m^2 = 3 + 3m^2\)

\( \implies 4m^2 – 3m^2 = 3\)

\(\implies m^2 = 3\)

\( \therefore m = \pm \sqrt3\)

\(\therefore\) Required equations are

\(\pm \sqrt 3x - y \mp \sqrt3 = 0\)

i.e., \(\sqrt3x - y - \sqrt3 =0=0 \)  \(and \,-\sqrt3x - y + \sqrt3 = 0 \)

⇒ \(\sqrt3x + y - \sqrt3 = 0\)

Given equation are y = mx + \(c_1\) …(i)

and y = mx + \(c_2\) …(ii)

Slopes of eq. (i) and eq. (ii) are same i.e., m So, they are parallel lines.

\(\therefore\) Distance between the two lines \(=\frac{|c_1 - c_2|}{\sqrt{1 + m^2}}\)

Given equation is y = 3x + 4

⇒ 3x – y + 4 = 0 …(i)

Slope = 3

Equation of any line passing through the point (2, 3) is

y – 3 = m(x – 2) …(ii)

If eq. (i) is perpendicular to eq. (ii) we get

m × 3 = - 1 \([\because m_1 × m_2 = - 1]\)

⇒ m = \(-\frac{1}{3}\)

Putting the value of m in eq. (ii) we get

y - 3 = \(-\frac{1}{3}\)(x-2)

⇒ 3y – 9 = - x + 2

⇒ x + 3y = 11 …(iii)

Solving eq. (i) and eq. (iii) we get

3x – y = - 4 ⇒ y = 3x + 4 …(iv)

Putting the value of y in eq. (iii) we get

x + 3(3x + 4) = 11

⇒ x + 9x + 12 = 11

⇒ 10x = – 1 ⇒ x = \(\frac{-1}{10}\)

From eq. (iv) we get, y = 3\((\frac{-1}{10}) + 4\)

⇒ \(y = \frac{-3}{10} + 4 \implies y = \frac{37}{10}\)

So the required coordinates are \((\frac{-1}{10}, \frac{37}{10})\)

Given equation is y = 3x – 1

Slope = 3

Since, required line is parallel to the line y = 3x - 1.

Therefore, slope of required line is m = 3.

So, the equation of the required line is

y – 2 = 3(x – 1)

Different form of equation of straight line are Slope intercept form, y = mx + c, Parameter = 2

Intercept form, \(\frac{x}{a} + \frac{y}{b} = 1\), parameter = 2

One-point form, y – \(y_1 = m(x – x_1)\), Parameter = 2

Normal form, x cos w + y sin w = P, Parameter = 2

Let the reflection of A(4, 1) about y = x be B(a, b).

Mid-point of AB = \((\frac{4 + a}{2}, \frac{1 + b}{2})\) which lies on y = x

⇒ \(\frac{4 +a}{2} = \frac{1 +b}{2}\)

⇒ 4 + a = 1 + b

⇒ a – b = - 3 …(i)

The slope of the line y = x is 1 and slope of AB = \(\frac{b - 1}{a - 4}\)

\(\therefore 1(\frac{b-1}{a-4}) = -1\)

⇒ b – 1 = – a + 4

⇒ a + b = 5 …(ii)

Solving eq. (i) and eq. (ii) we get

a = 1 and b = 4

∴ The point after translation is (1 + 2, 4) or (3, 4).

Given equation are

4x + 3y + 10 = 0 …(i)

5x – 12y + 26 = 0 …(ii)

and 7x + 24y – 50 = 0 …(iii)

Let \( (x_1 , y_1)\) be any point equidistant from eq. (i), eq. (ii) and eq. (iii).

Distance of \( (x_1 , y_1)\) from eq. (i)

\(= |\frac{4x_1 + 3y_1 + 10}{\sqrt{16 + 9}}| = |\frac{4x_1 + 3y_1 + 10}{5}|\)

Distance of \( (x_1 , y_1)\) from eq. (ii)

\(= |\frac{5x_1 - 12y_1 + 26}{\sqrt{25 + 144}}| = |\frac{5x_1 - 12y_1 + 26}{13}|\)

Distance of \( (x_1 , y_1)\) from eq. (iii)

\(= |\frac{7x_1 + 24y_1 - 50}{\sqrt{49 + 576}}| = |\frac{7x_1 + 24y_1 - 50}{25}|\)

If the point \( (x_1 , y_1)\) is equidistant from the given lines, then

\(|\frac{4x_1 + 3y_1 + 10}{5}| = |\frac{5x_1 - 12y_1 + 26}{13}|\\ = |\frac{7x_1 + 24y_1 - 50}{25}|\)

We see that putting \(x_1\) = 0 and \(y_1\) = 0, the above relation is satisfied i.e.,

\(\frac{10}{5} = \frac{26}{13} = \frac{50}{25} = 2\)

Any line perpendicular to 3x + y = 3 is x – 3y = λ (λ = constant)

If it passes through the point (2, 2) then

2 – 3(2) = λ ⇒ λ = - 4

∴ Required equation is x – 3y = - 4

⇒ - 3y = - x – 4

⇒ y = \(\frac{1}{3}x + \frac{4}{3} [\because y = mx + c]\)

So, the y-intercept is \(\frac{4}{3}\)