Question

$\large{ \red\bigstar} \underline{\huge \bf{Question }}$
Find out:–
Value of a for which the straight line 2x + y + 5 + a (x + y) = 0 is,
i) parallel to x = a
ii) parallel to y = a
iii) perpendicular to line 3x + 4y = 5​

Answer 1

$\large \bf \underline{ \underline{Question : - - - }}$

❍ Find the value of a for which the straight line 2x + y + 5 + a (x + y) = 0 is

• i) parallel to x = a
• ii) parallel to y = a
• iii) perpendicular to line 3x + 4y = 5

$\underline{\rule{72mm}{1mm}}$

$\huge \underline{ \underline{ \bf\: Solution - - - }}$

Given equation of line is

$\qquad \qquad \sf ↠2x + y + 5 + a(x + y) = 0 \\ \: \bf \: or \\ \sf \qquad \qquad↠(2 + a)x + (1 + a)y + = 0$

$\\ \sf: ⟹ \: (1 + a)y = - (2 + a)x - 5 \\ \\ \\ \sf: ⟹y = - \frac{(2 + a)}{(1 + a)} \: x \: - \frac{5}{(1 + a)} \\ \\ \\ \\ \: \: \underline { \: \: \: \: \: \blue{∴ \: \: \textbf{Gradient of this line = } - \frac{2 + a}{1 + a} \: \: \: \: \: }}$

____________

$\\ \textbf{(i) Obviously, line x = a makes an angle 90° with X-axis}$

$\\ \bf \red{ ∴ It's \: gradient = tan90° = \infty }$

$\\ \bf \: ∵ Given \: line \: is \: parallel to \: this \: line \\ \\ \sf ∴ \: - \frac{2 + a}{1 + a} = \infty \\ \\ \\ \\ ⟹ \frac{1 + a}{2 + a} = 0 \\ \\ \\ \\ ⟹1 + a = 0 \\ \\ \\ \\ \pink{ \large{ ⟹\underline{ \boxed{ \bf a = - 1}}}}$

____________

$\\ \textbf{(ii) Obviously, line y = a makes an angle 0° with X-axis. }$

$\\∴\:\bf\red{ It's \:Gradient = tan0° = 0}$

$\\ ∵ \bf \: It \: is \: parallel \: t o \: given \: line \\\\∴ \sf \: - \frac{2 + a}{1 + a} = 0 \\ \\ \\ \\ ⟹ \sf \: 2 + a = 0 \\ \\ \\ \\ \large \pink{⟹ \underline{ \boxed{ \bf \: a = - 2}}}$

____________

$\\ \textbf{(iii)Here ,equation of the line is :}$

$\\ \\ \qquad \qquad \: \bf \: 3x + 4y = 5 \\ \\ \\ \sf \: ⟹4y = 5 - 3x \\ \\ \\ \sf⟹y = - \frac{3}{4} x + \frac{5}{4} \\ \\ \\ \qquad \qquad\large \underline { \pink{ \bf \: ∴It's \: gradient \: = - \frac{3}{4} } \: \: \: }$

$\\ \textbf {∵ This is perpendicular to given line }$

$\\ ∴\qquad \qquad \: \bf \: m_1 × m_2 = - 1$

$\\ \sf⟹ \bigg( - \frac{2 + a}{1 + a} \bigg) \times \bigg( - \frac{ 3 }{4} \bigg) = - 1 \\ \\ \\ \\ \sf ⟹ \frac{3(2 + a)}{4(1 + a)} = - 1 \\ \\ \\ ⟹ \sf \frac{6 + 3a}{4 + 4a} = - 1 \\ \\ \\ ⟹ \sf6 + 3a = - 4 - 4a \\ \\ \\ \sf ⟹7a = - 10 \\ \\ \\ \qquad \underline{\large \pink{⟹ \boxed{ \bf \: a = - \frac{10}{7} }} \qquad}$

Answer 2

Answer:

Find the value of a for which the straight line 2x + y + 5 + a (x + y) = 0 is

i) parallel to x = a

ii) parallel to y = a

iii) perpendicular to line 3x + 4y = 5

\underline{\rule{72mm}{1mm}}

\huge \underline{ \underline{ \bf\: Solution - - - }}

Solution−−−

Given equation of line is

\begin{gathered} \qquad \qquad \sf ↠2x + y + 5 + a(x + y) = 0 \\ \: \bf \: or \\ \sf \qquad \qquad↠(2 + a)x + (1 + a)y + = 0\end{gathered}

↠2x+y+5+a(x+y)=0

or

↠(2+a)x+(1+a)y+=0

\begin{gathered} \\ \sf: ⟹ \: (1 + a)y = - (2 + a)x - 5 \\ \\ \\ \sf: ⟹y = - \frac{(2 + a)}{(1 + a)} \: x \: - \frac{5}{(1 + a)} \\ \\ \\ \\ \: \: \underline { \: \: \: \: \: \blue{∴ \: \: \textbf{Gradient of this line = } - \frac{2 + a}{1 + a} \: \: \: \: \: }}\\\\\end{gathered}

:⟹(1+a)y=−(2+a)x−5

:⟹y=−

(1+a)

(2+a)

x−

(1+a)

5

∴Gradient of this line = −

1+a

2+a

____________

\begin{gathered} \\ \textbf{(i) Obviously, line x = a makes an angle 90° with X-axis}\end{gathered}

(i) Obviously, line x = a makes an angle 90° with X-axis

\begin{gathered} \\ \bf \red{ ∴ It's \: gradient = tan90° = \infty } \\ \\ \end{gathered}

∴It

′

sgradient=tan90°=∞

\begin{gathered} \\ \bf \: ∵ Given \: line \: is \: parallel to \: this \: line \\ \\ \sf ∴ \: - \frac{2 + a}{1 + a} = \infty \\ \\ \\ \\ ⟹ \frac{1 + a}{2 + a} = 0 \\ \\ \\ \\ ⟹1 + a = 0 \\ \\ \\ \\ \pink{ \large{ ⟹\underline{ \boxed{ \bf a = - 1}}}} \\ \\ \end{gathered}