मराठी

The Equation of the Line with Slope −3/2 and Which is Concurrent with the Lines 4x + 3y − 7 = 0 and 8x + 5y − 1 = 0 is - Mathematics

Advertisements
Advertisements

प्रश्न

The equation of the line with slope −3/2 and which is concurrent with the lines 4x + 3y − 7 = 0 and 8x + 5y − 1 = 0 is

पर्याय

  •  3x + 2y − 63 = 0

  •  3x + 2y − 2 = 0

  • 2y − 3x − 2 = 0

  • none of these

MCQ
Advertisements

उत्तर

 3x + 2y − 2 = 0

Given:
4x + 3y − 7 = 0      ... (1)
8x + 5y − 1 = 0      ... (2)
The equation of the line with slope \[- \frac{3}{2}\] is given below: \[y = - \frac{3}{2}x + c\] \[\Rightarrow \frac{3}{2}x + y - c = 0\]          ... (3)
The lines (1), (2) and (3) are concurrent.

\[\therefore \begin{vmatrix}4 & 3 & - 7 \\ 8 & 5 & - 1 \\ \frac{3}{2} & 1 & - c\end{vmatrix} = 0\]

\[ \Rightarrow 4\left( - 5c + 1 \right) - 3\left( - 8c + \frac{3}{2} \right) - 7\left( 8 - \frac{15}{2} \right) = 0\]

\[ \Rightarrow - 20c + 4 + 24c - \frac{9}{2} - 56 + \frac{105}{2} = 0\]

\[ \Rightarrow \frac{- 40c + 8 + 48c - 9 - 112 + 105}{2} = 0\]

\[ \Rightarrow 8c = 8\]

\[ \Rightarrow c = 1\]

On substituting c = 1 in \[y = - \frac{3}{2}x + c\], we get:

\[y = - \frac{3}{2}x + 1\]

\[ \Rightarrow 3x + 2y - 2 = 0\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 23: The straight lines - Exercise 23.21 [पृष्ठ १३५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 23 The straight lines
Exercise 23.21 | Q 30 | पृष्ठ १३५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

The base of an equilateral triangle with side 2a lies along they y-axis such that the mid point of the base is at the origin. Find vertices of the triangle.


Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, –4) and B (8, 0).


The slope of a line is double of the slope of another line. If tangent of the angle between them is `1/3`, find the slopes of the lines.


If three point (h, 0), (a, b) and (0, k) lie on a line, show that `q/h + b/k = 1`


Find the values of k for which the line (k–3) x – (4 – k2) y + k2 –7k + 6 = 0 is 

  1. Parallel to the x-axis,
  2. Parallel to the y-axis,
  3. Passing through the origin.

Find the slope of a line passing through the following point:

\[(a t_1^2 , 2 a t_1 ) \text { and } (a t_2^2 , 2 a t_2 )\]


State whether the two lines in each of the following are parallel, perpendicular or neither.

Through (9, 5) and (−1, 1); through (3, −5) and (8, −3)


Using the method of slope, show that the following points are collinear A (4, 8), B (5, 12), C (9, 28).


Using the method of slope, show that the following points are collinear A (16, − 18), B (3, −6), C (−10, 6) .


What can be said regarding a line if its slope is  zero ?


If three points A (h, 0), P (a, b) and B (0, k) lie on a line, show that: \[\frac{a}{h} + \frac{b}{k} = 1\].


Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear.


By using the concept of slope, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are the vertices of a parallelogram.


Find the equation of a straight line with slope 2 and y-intercept 3 .


Show that the perpendicular bisectors of the sides of a triangle are concurrent.


Find the equations of the altitudes of a ∆ ABC whose vertices are A (1, 4), B (−3, 2) and C (−5, −3).


Find the acute angle between the lines 2x − y + 3 = 0 and x + y + 2 = 0.


Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.


Find the tangent of the angle between the lines which have intercepts 3, 4 and 1, 8 on the axes respectively.


Show that the tangent of an angle between the lines \[\frac{x}{a} + \frac{y}{b} = 1 \text { and } \frac{x}{a} - \frac{y}{b} = 1\text {  is } \frac{2ab}{a^2 - b^2}\].


The acute angle between the medians drawn from the acute angles of a right angled isosceles triangle is 


The medians AD and BE of a triangle with vertices A (0, b), B (0, 0) and C (a, 0) are perpendicular to each other, if


The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y − 11 = 0 are


Point of the curve y2 = 3(x – 2) at which the normal is parallel to the line 2y + 4x + 5 = 0 is ______.


A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and then passes through the point (5, 3). Find the coordinates of the point A.


The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 are ______.


Show that the tangent of an angle between the lines `x/a + y/b` = 1 and `x/a - y/b` = 1 is `(2ab)/(a^2 - b^2)`


P1, P2 are points on either of the two lines `- sqrt(3) |x|` = 2 at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P1, P2 on the bisector of the angle between the given lines.


If p is the length of perpendicular from the origin on the line `x/a + y/b` = 1 and a2, p2, b2 are in A.P, then show that a4 + b4 = 0.


The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is ______.


The points (3, 4) and (2, – 6) are situated on the ______ of the line 3x – 4y – 8 = 0.


The line `x/a + y/b` = 1 moves in such a way that `1/a^2 + 1/b^2 = 1/c^2`, where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is x2 + y2 = c2.


Line joining the points (3, – 4) and (– 2, 6) is perpendicular to the line joining the points (–3, 6) and (9, –18).


The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and

Column C1 Column C2
(a) Through the point (2, 1) is (i) 2x – y = 4
(b) Perpendicular to the line (ii) x + y – 5
= 0 x + 2y + 1 = 0 is
(ii) x + y – 5 = 0
(c) Parallel to the line (iii) x – y –1 = 0
3x – 4y + 5 = 0 is
(iii) x – y –1 = 0
(d) Equally inclined to the axes is (iv) 3x – 4y – 1 = 0

A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and then passes through the point (5, 3). The co-ordinates of the point A is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×