Question

The vertices of a triangle OBC are 0 (0,0), B (-3,-1), C (-1-3). Find the equation of the line parallel to BC and intersecting the sides OB and OC and whose perpendicular distance from the origin is 1/2​

Answer 1

Step-by-step explanation:

Given The vertices of a triangle OBC are 0 (0,0), B (-3,-1), C (-1-3). Find the equation of the line parallel to BC and intersecting the sides OB and OC and whose perpendicular distance from the origin is 1/2

• Now we have a triangle OBC. Draw a line parallel to BC and let it be PQ. So we need to find the equation for PQ.
• The distance of the perpendicular line is ½ .
• So equation of line BC will be y – y1 = y2 – y1 / x2 – x1 (x – x1)
• Taking the points we have    y + 1 = - 3 + 1 / - 1 + 3 (x + 3)
•                                            So y + 1 = - 1 (x + 3)
•                                            So y + 1 = - x – 3
•                                              So y = -x – 4
• Now equation of PQ will be y = mx + c
•           Now BC is parallel to PQ
• So slope of line BC = slope of line PQ
• Slope of line PQ = - 1
•           So y = mx + c
•         Or y = -- x + c
• Distance is (0,0)
• Taking equation y + x – c = 0
• So we have mod - c / √1^2 + 1^2 = ½ will be the distance from origin
• Therefore c / √2 = ½
•               Or c = √2 / 2
•                Or c = 1 / √2
• So equation of PQ will be y = - x + 1/√2
•                                    Or √2 y = - √2 x + 1
•                                  Or √2 x + √2 y – 1 = 0

Reference link will be

https://brainly.in/question/17309836

Answer 2

Answer:

√2x+√2y-1=0

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