<prompt> Fumana equation ea mola o fetang ntlha e fanoeng me o bapile le vector e fanoeng.Ntlha: 2, 1, Vector: 2, 1  <response>: Instruction in English: Find the equation of the line which passes through the given point and is parallel to the given vector. . Response in English: Lets let the equation of the line be expressed in the form y mx b, where m is the slope of the line and b is the y-intercept. First, we can find the slope by observing that the slope of the line parallel to the vector 2,1 is the ratio of the y-component to the x-component of the vector. This gives us a slope of 1-2 -12. Now, we can use the point-slope form of the line equation: y - y1 mx - x1, where x1, y1 is the given point 2,1.Substituting the values, we get: y - 1 -12x - 2.Finally, we can put this equation into slope intercept form by distributin -12 and solving for y:y - 1 -12x - 2y - 1 -12 x 1y -12 x 2Thus, the equation of the line that passes through the point 2,1 and is parallel to the vector 2,1 is y -12 x 2. . Response in Sesotho: re lumelle hore equation ea mola e hlahisoe ka mokhoa oa y mx b, moo m e leng letsoapo la mola me b e le y-intercept. Ntlha ea pele, re ka fumana moepa ka ho hlokomela hore moepa oa mola o bapileng le vector -2,1 ke karo-karolelano ea karolo ea y ho ea x-karolo ea vector. Sena se re fa moepa oa 1-2 -12. Hona joale, re ka sebelisa sebopeho sa moepa oa mola oa equation: y - y1 mx - x1, moo x1, y1 e leng ntlha e fanoeng 2,1. Ho fetola litekanyetso, re fumana: y - 1 -12x - 2. Qetellong, re ka beha equation ena ka mokhoa oa ho thibela letsoapo ka ho aba -12 le ho rarolla y: y - 1 -12x - 2 y - 1 -12 x 1 y -1 2 x 2 Kahoo, equation ea mola o fetang ntlha 2,1 mme o bapile le vector -2,1 ke y -12 x 2. 