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In Class 11, Chapter 23 of RD Sharma's textbook focuses on the concept of straight lines which is a fundamental topic in coordinate geometry. Exercise 23.3 deals with advanced problems involving the properties and equations of straight lines. This exercise helps students understand the practical applications of the line equations and geometric relationships between the different lines.
The Straight lines in coordinate geometry are described by their equations which can be used to find various properties such as the slope, intercepts, and distance between the lines. The general form of a line equation is Ax+By+C=0. Through various problems, students explore concepts such as the parallel and perpendicular lines distance from the point to a line and intersections. Mastery of these concepts is crucial for solving complex geometric problems and for applications in higher mathematics.
Solution:
Given, slope, m = tan (150°) ⇒ m = -1/√3, also, y-intercept = (0,2)
We know that equation, of a line is given as y = mx+ c, m is the slope and c is the intercept that line cuts on y-axis, therefore
equation for the line will be:
y = -x/√3 + 2
⇒ x - 2√3 + √3y = 0
(i) with slope 2 and y-intercept 3;
(ii) with slope -1/3 and y-intercept -4
(iii) with slope -2 and intersecting the x-axis at a distance 3 units to the left of the origin.
Solution:
(i) We know that equation, of a line is given as y = mx+ c, therefore equation for the line with slope 2 and y-intercept 3 will be: y = 2x + 3
(ii) Similarly, equation of the line with slope -1/3 and y-intercept -4 will be: y = -x/3 - 4
⇒ x +3y +12 = 0
(iii) Since, the line cuts the x-axis at a distance 3 units to the left of origin its coordinate will be (-3,0) and the given slope, m = -2.
Equation of a line passing through a point is given by the formula: y-y1 = m (x - x1), hence the equation will be
⇒ y - 0 = -2(x - (-3))
⇒ y = -2x -6
⇒ 2x + y + 6 = 0
Solution:
The equation of the line on the coordinate axes are x=0 and y=0.
The equations of the bisectors of the angle between x=0 and y=0 are:
x ± y = 0
Solution:
Here, ∅ = tan-1 (3) ⇒ m = tan∅ =3, line cuts cuts off an intercept of 4 units on negative direction of y-axis, so the coordinate will be (0, -4)
Hence, the equation of the line is: y = 3x -4
Solution:
Since, our required line is parallel to the line passing through the coordinates (2, -5) and (1,2), it will have the same slope as the later line. Therefore, slope, m = (y2 - y1) / (x2 - x1) = (2 - (-5) ) / (1 - 2 ) = -7
Also, given y-intercept = -4, hence the required equation of the line is: y = -7x - 4
Solution:
Slope of the line passing through the points (4,2) and (3,5) is
(y2 - y1) / (x2 - x1) = (5 - 2 ) / (3 - 4 ) = -3
Now, our required line is perpendicular to the former line, so its slope will be m = 1/3
Also, y-intercept, c = 3, hence the required equation of the line is: y = x/3 + 3
⇒ x - 3y +9 = 0
Solution:
Slope of the line passing through the points (4,3) and (-1,1) is
(y2 - y1) / (x2 - x1) = (1 - 3 ) / (-1 - 4 ) = 2/5
Now, our required line is perpendicular to the former line, so its slope will be m = -5/2
Also, y-intercept, c = -3, hence the required equation of the line is: y = -5x/2 - 3
⇒ 5x + 2y +6 = 0
Solution:
Given, m = tan 30° = 1/√3
Since, the line intersects the y-axis at a distance 2 units above the origin its coordinate will be (0,2)
Equation of a line passing through a point is given by the formula: y-y1 = m (x - x1), hence the equation will be
⇒ y - 2 = 1/√3 . (x - 0)
⇒ √3y - 2√3 = x
⇒ x - √3y + 2√3 = 0
Read More:
Exercise 23.3 of Chapter 23 provides the rigorous practice for the solving problems related to the straight lines. By working through these exercises students gain a deeper understanding of the line equations and their properties which are essential for both the theoretical and applied mathematics. Consistent practice of these problems enhances problem-solving skills and prepares students for the more advanced topics in the coordinate geometry.
