Solution: Given line equation: 4x – 3y – 5 = 0 3y = 4x – 5 y = (4/3) x – 5 Slope of the line (m1) = 4/3 Let the slope of the line perpendicular to the given line be m2 Then, m1 x m2 = -1 (4/3) x...
Solution: Let’s assume the point of intersection of the line 2x + 5y – 4 = 0 and x-axis be (x, 0) Now, substituting the value y = 0 in the line equation, we have 2x + 5(0) – 4 = 0 2x – 4 = 0 x = 4/2...
Solution: Given, co-ordinates of points E and F are (0, 4) and (3, 7) respectively (i) The gradient of EF m = y2 – y1 / x2 – x1 = (7 – 4)/(3 – 0) = 3/3 ⇒ m = 1 (ii) Equation of line EF is given by,...
Solution: Given, Inclination of a straight line is 60o So, the slope = tan 60o = √3 = m And, the equation of line passes through the point (0, -3) = (x1, y1) Hence, the equation of line is given by...
Solution: Given, (i) tan θ = 1 ⇒ θ = 45o (ii) tan θ = √3 ⇒ θ = 60o
Solution: The slope of a line having inclination: (i) 45o Slope = tan 45o = 1 (ii) 30o Slope = tan 30o = 1/√3