US Corrected Text
Slide 1: Condition for Perpendicular Lines in Terms of Their Slopes
Learning Objective
At the end of this class session, you will have well understood about the condition for perpendicular lines in terms of their slopes.
Slide 2: Introduction
Slope is the ratio of the change in the horizontal distance to the change in the vertical distance.
Slide 3: Explanation_1
Area of a Quadrilateral
Teacher: Hi students, today we are going to discuss about the condition for perpendicular lines in terms of their slopes.
Student: Okay, mam.
Teacher: Two lines are said to be perpendicular if the angle between them is exactly equal to 90 degrees.
Student: Okay, mam.
Teacher: Two lines are said to be perpendicular if the product of …show more content…
The slope of a straight line joining the points (x2, y2) and (x3, y3) is m2= (y_3- y_2)/(x_3- x_2 ). Since n1 and n2 are perpendicular straight lines they form a right angle at R. {where is C? - sar} Therefore, PQR forms a right triangle. Thus, we can apply the Pythagoras theorem that PQ2 = PR2 + RQ2. Applying the distance formula we get, (x2 - x1)2 + (y2 – y1)2 = (x3 – x1)2 + (y3 – y1)2 + (x3 – x2)2+(y3 – y2)2. Add and subtract x3 and y3 to the LHS. Thus, the previous step becomes, (x2 - x3 + x3 - x1)2 + (y2 – y3 + y3- y1)2 = (x3 – x1)2 + (y3 – y1)2 + (x3 – x2)2+(y3 – y2)2. By grouping and squaring we get, (x2 – x3)2 + (x3 – x1)2 + 2(x2 – x3)(x3 – x1) + (y2 – y3)2 + (y3 – y1)2 + 2(y2 – y2)(y3 - y1) = (x3 – x1)2 + (y3 – y1)2 + ( x3 – x2)2 + (y3 – y2)2. By cancelling the common terms we obtain, 2(x2 – x3)(x3 – x1) + 2(y2 – y3)(y3 – y1) = 0. It implies (y2 – y3)(y3 – y1) = -(x2 – x3)(x3 – x1)→((y_3-y_1)/(x_3- x_1 ))((y_3- y_2)/(x_3- x_2 ))= -1 We know that m1 = (y_3-y_1)/(x_3- x_1 ) and m2 = (y_3- y_2)/(x_3- x_2 ); substituting these values we get, m1m2 = -1 Hence, m1 = (-1)/m_2 is the condition for two straight lines to be
Slide 1: Condition for Perpendicular Lines in Terms of Their Slopes
Learning Objective
At the end of this class session, you will have well understood about the condition for perpendicular lines in terms of their slopes.
Slide 2: Introduction
Slope is the ratio of the change in the horizontal distance to the change in the vertical distance.
Slide 3: Explanation_1
Area of a Quadrilateral
Teacher: Hi students, today we are going to discuss about the condition for perpendicular lines in terms of their slopes.
Student: Okay, mam.
Teacher: Two lines are said to be perpendicular if the angle between them is exactly equal to 90 degrees.
Student: Okay, mam.
Teacher: Two lines are said to be perpendicular if the product of …show more content…
The slope of a straight line joining the points (x2, y2) and (x3, y3) is m2= (y_3- y_2)/(x_3- x_2 ). Since n1 and n2 are perpendicular straight lines they form a right angle at R. {where is C? - sar} Therefore, PQR forms a right triangle. Thus, we can apply the Pythagoras theorem that PQ2 = PR2 + RQ2. Applying the distance formula we get, (x2 - x1)2 + (y2 – y1)2 = (x3 – x1)2 + (y3 – y1)2 + (x3 – x2)2+(y3 – y2)2. Add and subtract x3 and y3 to the LHS. Thus, the previous step becomes, (x2 - x3 + x3 - x1)2 + (y2 – y3 + y3- y1)2 = (x3 – x1)2 + (y3 – y1)2 + (x3 – x2)2+(y3 – y2)2. By grouping and squaring we get, (x2 – x3)2 + (x3 – x1)2 + 2(x2 – x3)(x3 – x1) + (y2 – y3)2 + (y3 – y1)2 + 2(y2 – y2)(y3 - y1) = (x3 – x1)2 + (y3 – y1)2 + ( x3 – x2)2 + (y3 – y2)2. By cancelling the common terms we obtain, 2(x2 – x3)(x3 – x1) + 2(y2 – y3)(y3 – y1) = 0. It implies (y2 – y3)(y3 – y1) = -(x2 – x3)(x3 – x1)→((y_3-y_1)/(x_3- x_1 ))((y_3- y_2)/(x_3- x_2 ))= -1 We know that m1 = (y_3-y_1)/(x_3- x_1 ) and m2 = (y_3- y_2)/(x_3- x_2 ); substituting these values we get, m1m2 = -1 Hence, m1 = (-1)/m_2 is the condition for two straight lines to be