COORDINATE GEOMETRY ( ONLINE TUITION) {+1, GEOMETRY (HON), PRE-CAL}

COORDINATE GEOMETRY ( ONLINE TUITION) ============================================ RELEVANT CLASSES / COURSES : Algebra 1 & 2 , Geometry (Hon) , Pre- Calculus , CBSE + 1 , ISC + 1 , A & AS Level (UK), IB ............... Abbreviations used in this paper Abs: absolute value sqrt: square root ======================================================= Question 1 Find the equation of the line perpendicular to the line x – 2y + 3 = 0 and passing through the point ( 1 , - 2 ) Help for solution The gradient/slope of the given line x – 2y + 3 = 0 will be ½ Hence slope of the perpendicular will be - 2 The line passes through ( 1 , - 2 ) [ Given] Now find the equation of the required line applying (y - y1) = m(x - x1) Answer : y + 2x = 0 ====================================================== Question 2 Find the distance from the point ( - 2 , - 1 ) to the line 4x + 3y = 5 Help for solution Write down the given line the general form : Ax1+ By1+ c = 0 Then we get , 4x + 3y – 5 = 0 Then apply the formula , distance, d = Abs [ (Ax1 + By1+ C)/sqrt(A2+B2)] Answer : d = 16/5 units ======================================================= Question 3 The vertices of a triangle are A(-2 , 1) , B(2 , 3) and C (-2 , - 4). Find the value of tan B Help for solution Slope of AB = m1 = ½ (Find it yourself !) Slope of BC = m2 = 7/4 (Find it yourself) Now use the formula tan B = (m2 – m1) / (1 + m1m2) Answer : tan B = 2/3 ====================================================== Question 4 The sum of the intercepts made by a line on the coordinate axes is 9 and this line passes through (2 , 2) Find the equation of the line . Help for solution Assume the intercepts as , a and (9 – a) Then the equation of the required line : x/a + y/ (9 - a) = 1 Substitute x = 2 and y = 2 in the above , simplify it and obtain the equation, a2 - 9a + 18 = 0 Solve the above equation and find the possible x and corresponding y intercepts. They will be a = 3 , b= 6 OR a = 6 , b = 3 Now obtain the two possible equations of the required line by substituting for the values of a and b in the intercept form of the equation, x/a + y/b = 1 Answer : The simplified equations will be x + 2y - 6 = 0 and 2x + y – 6 = 0 ======================================================= Question 5 Using the concept of gradient/slope , prove that the points A ( 3, 0) , B (-2 , -2) and C ( 8 , 2) are collinear Help for solution Find the slope/gradient of AB, (m1) using the formula, m = (y2 - y1) / (x2 - x1) Then find the slope of BC , m2 The above slopes m1 and m2 will be found as equal Answer : So the three points lie on the same line or collinear ======================================================= Question 6 Find the equation of the perpendicular bisector of the line joining A (-2 , 5 ) to B(-8 , -3) Help for solution Mind the midpoint of AB using midpoint formula [Ans: (-5 , 1] Find the gradient/slope of AB [Ans: 4/3] So slope of a line perpendicular to AB = -3/4 Apply y - y1 = m(x - x1) to find the required equation ie y – 1 = -3/4( x - (-5)) Answer : 4y + 3x + 11 = 0 ====================================================