RD Chapter 23 The Straight Lines Ex 23.19 Solutions
Question - 1 : - Find the equation of a straight line through the point of intersection of the lines 4x тАУ 3y = 0 and 2x тАУ 5y + 3 = 0 and parallel to 4x + 5 y + 6 = 0.
Answer - 1 : -
Given:
Lines 4x тАУ 3y = 0 and2x тАУ 5y + 3 = 0 and parallel to 4x + 5 y + 6 = 0
The equation of thestraight line passing through the points of intersection of 4x┬атИТ┬а3y =0 and 2x┬атИТ┬а5y + 3 = 0 is given below:
4x┬атИТ┬а3y+┬а╬╗┬а(2x┬атИТ┬а5y + 3) = 0
(4 + 2╬╗)x +(тИТ┬а3┬атИТ┬а5╬╗)y + 3╬╗┬а= 0
The required line isparallel to 4x + 5y + 6 = 0 or, y = -4x/5 тАУ 6/5
╬╗ = -16/15
тИ┤ The required equationis
28x + 35y тАУ 48 = 0
Question - 2 : - Find the equation of a straight line passing through the point of intersection of x + 2y + 3 = 0 and 3x + 4y + 7 = 0 and perpendicular to the straight line x тАУ y + 9 = 0.
Answer - 2 : -
Given:
x + 2y + 3 = 0 and 3x+ 4y + 7 = 0
The equation of thestraight line passing through the points of intersection of x + 2y + 3 = 0 and3x + 4y + 7 = 0 is
x + 2y + 3 +┬а╬╗(3x+ 4y + 7) = 0
(1 + 3╬╗)x + (2 + 4╬╗)y+ 3 + 7╬╗┬а= 0
The required line isperpendicular to x┬атИТ┬аy + 9 = 0 or, y = x + 9
Question - 3 : - Find the equation of the line passing through the point of intersection of 2x тАУ 7y + 11 = 0 and x + 3y тАУ 8 = 0 and is parallel to (i) x = axis (ii) y-axis.
Answer - 3 : -
Given:
The equations, 2x тАУ 7y+ 11 = 0 and x + 3y тАУ 8 = 0
The equation of thestraight line passing through the points of intersection of 2x┬атИТ┬а7y +11 = 0 and x + 3y┬атИТ┬а8 = 0 is given below:
2x┬атИТ┬а7y + 11+┬а╬╗(x + 3y┬атИТ┬а8) = 0
(2 +┬а╬╗)x +(тИТ┬а7 + 3╬╗)y + 11┬атИТ┬а8╬╗┬а= 0
(i)┬аThe requiredline is parallel to the x-axis. So, the coefficient of x should be zero.
2 +┬а╬╗┬а= 0
╬╗┬а= -2
Now, substitute thevalue of ╬╗ back in equation, we get
0 +(тИТ┬а7┬атИТ┬а6)y + 11 + 16 = 0
13y┬атИТ┬а27 = 0
тИ┤ The equation of therequired line is 13y┬атИТ┬а27 = 0
(ii)┬аThe requiredline is parallel to the y-axis. So, the coefficient of y should be zero.
-7 + 3╬╗┬а= 0
╬╗ = 7/3
Now, substitute thevalue of ╬╗ back in equation, we get
(2 + 7/3)x + 0 + 11 тАУ8(7/3) = 0
13x тАУ 23 = 0
тИ┤ The equation of therequired line is 13x тАУ 23 = 0
Question - 4 : - Find the equation of the straight line passing through the point of intersection of 2x + 3y + 1 = 0 and 3x тАУ 5y тАУ 5 = 0 and equally inclined to the axes.
Answer - 4 : -
Given:
The equations, 2x + 3y+ 1 = 0 and 3x тАУ 5y тАУ 5 = 0
The equation of thestraight line passing through the points of intersection of 2x + 3y + 1 = 0 and3x┬атИТ┬а5y┬атИТ┬а5 = 0 is
2x + 3y + 1+┬а╬╗(3x┬атИТ┬а5y┬атИТ┬а5) = 0
(2 + 3╬╗)x +(3┬атИТ┬а5╬╗)y + 1┬атИТ┬а5╬╗┬а= 0
y = тАУ [(2 + 3╬╗) / (3 тАУ5╬╗)] тАУ [(1 тАУ 5╬╗) / (3 тАУ 5╬╗)]
The required line isequally inclined to the axes. So, the slope of the required line is either 1or┬атИТ┬а1.
So,
тАУ [(2 + 3╬╗) / (3 тАУ5╬╗)] = 1 and тАУ [(2 + 3╬╗) / (3 тАУ 5╬╗)] = -1
-2 тАУ 3╬╗┬а= 3 тАУ5╬╗┬аand 2 + 3╬╗┬а= 3 тАУ 5╬╗
╬╗ = 5/2 and 1/8
Now, substitute thevalues of┬а╬╗┬аin (2 + 3╬╗)x + (3┬атИТ┬а5╬╗)y +1┬атИТ┬а5╬╗┬а= 0, we get the equations of the required lines as:
(2 + 15/2)x + (3 тАУ25/2)y + 1 тАУ 25/2 = 0 and (2 + 3/8)x + (3 тАУ 5/8)y + 1 тАУ 5/8 = 0
19x тАУ 19y тАУ 23 = 0 and19x + 19y + 3 = 0
тИ┤ The required equationis 19x тАУ 19y тАУ 23 = 0 and 19x + 19y + 3 = 0
Question - 5 : - Find the equation of the straight line drawn through the point of intersection of the lines x + y = 4 and 2x тАУ 3y = 1 and perpendicular to the line cutting off intercepts 5, 6 on the axes.
Answer - 5 : -
Given:
The lines x + y = 4and 2x тАУ 3y = 1
The equation of thestraight line passing through the point of intersection of x + y = 4 and2x┬атИТ┬а3y = 1 is
x + y┬атИТ┬а4+┬а╬╗(2x┬атИТ┬а3y┬атИТ┬а1) = 0
(1 + 2╬╗)x +(1┬атИТ┬а3╬╗)y┬атИТ┬а4┬атИТ┬а╬╗┬а= 0 тАж (1)
y = тАУ [(1 + 2╬╗) / (1 тАУ3╬╗)]x + [(4 + ╬╗) / (1 тАУ 3╬╗)]
The equation of theline with intercepts 5 and 6 on the axis is
x/5 + y/6 = 1 тАж. (2)
So, the slope of thisline is -6/5
The lines (1) and (2)are perpendicular.
тИ┤ -6/5 ├Ч [(-1+2╬╗) / (1тАУ 3╬╗)] = -1
╬╗ = 11/3
Now, substitute thevalues of┬а╬╗┬аin (1), we get the equation of the required line.
(1 + 2(11/3))x +(1┬атАУ┬а3(11/3))y┬атИТ┬а4┬атАУ┬а11/3┬а= 0
(1 + 22/3)x + (1 тАУ11)y тАУ 4 тАУ 11/3 = 0
25x тАУ 30y тАУ 23 = 0
тИ┤ The required equationis 25x тАУ 30y тАУ 23 = 0