Mathematics Past Paper PDF - Straight Lines - 2021-22

212 MATHEMATICS 6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle. 7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise. 8. Find the...

212 MATHEMATICS 6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle. 7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise. 8. Find the value of x for which the points (x, – 1), (2,1) and (4, 5) are collinear. 9. Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram. 10. Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2). 11. The slope of a line is double of the slope of another line. If tangent of the angle 1 between them is , find the slopes of the lines. 3 12. A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k – y1 = m (h – x1). a b 13. If three points (h, 0), (a, b) and (0, k) lie on a line, show that + = 1. h k 14. Consider the following population and year graph (Fig 10.10), find the slope of the line AB and using it, find what will be the population in the year 2010? Fig 10.10 10.3 Various Forms of the Equation of a Line We know that every line in a plane contains infinitely many points on it. This relationship between line and points leads us to find the solution of the following problem: 2021-22  STRAIGHT LINES 213 How can we say that a given point lies on the given line? Its answer may be that for a given line we should have a definite condition on the points lying on the line. Suppose P (x, y) is an arbitrary point in the XY-plane and L is the given line. For the equation of L, we wish to construct a statement or condition for the point P that is true, when P is on L, otherwise false. Of course the statement is merely an algebraic equation involving the variables x and y. Now, we will discuss the equation of a line under different conditions. 10.3.1 Horizontal and vertical lines If a horizontal line L is at a distance a from the x-axis then ordinate of every point lying on the line is either a or – a [Fig 10.11 (a)]. Therefore, equation of the line L is either y = a or y = – a. Choice of sign will depend upon the position of the line according as the line is above or below the y-axis. Similarly, the equation of a vertical line at a distance b from the y-axis is either x = b or x = – b [Fig 10.11(b)]. Fig 10.11 Example 6 Find the equations of the lines parallel to axes and passing through (– 2, 3). Solution Position of the lines is shown in the Fig 10.12. The y-coordinate of every point on the line parallel to x-axis is 3, therefore, equation of the line parallel tox-axis and passing through (– 2, 3) is y = 3. Similarly, equation of the line parallel to y-axis and passing through (– 2, 3) is x = – 2. Fig 10.12 2021-22 214 MATHEMATICS 10.3.2 Point-slope form Suppose that P0 (x0, y0) is a fixed point on a non-vertical line L, whose slope is m. Let P (x, y) be an arbitrary point on L (Fig 10.13). Then, by the definition, the slope of L is given by y − y0 m= , i.e. , y − y 0 = m (x − x 0 ) x − x0...(1) Since the point P0 (x0 , y0) along with all points (x, y) on L satisfies (1) and no Fig 10.13 other point in the plane satisfies (1). Equation (1) is indeed the equation for the given line L. Thus, the point (x, y) lies on the line with slope m through the fixed point (x0, y0), if and only if, its coordinates satisfy the equation y – y0 = m (x – x0) Example 7 Find the equation of the line through (– 2, 3) with slope – 4. Solution Here m = – 4 and given point (x0 , y0) is (– 2, 3). By slope-intercept form formula (1) above, equation of the given line is y – 3 = – 4 (x + 2) or 4x + y + 5 = 0, which is the required equation. 10.3.3 Two-point form Let the line L passes through two given points P1 (x1, y1) and P2 (x2, y2). Let P (x, y) be a general point on L (Fig 10.14). The three points P1, P2 and P are Fig 10.14 collinear, therefore, we have slope of P1P = slope of P1P2 y − y1 y 2 − y1 y 2 − y1 i.e., = , or y − y 1 = ( x − x1 ). x − x1 x 2 − x1 x 2 − x1 2021-22